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• Improving performance on data pasting 2000 rows with validations

  - by Lohit

  I have N rows (which could be nothing less than 1000) on an excel spreadsheet. And in this sheet our project has 150 columns like this: Now, our application needs data to be copied (using normal Ctrl+C) and pasted (using Ctrl+V) from the excel file sheet on our GUI sheet. Copy pasting 1000 records takes around 5-6 seconds which is okay for our requirement, but the problem is when we need to make sure the data entered is valid. So we have to validate data in each row generate appropriate error messages and format the data as per requirement. So we need to at runtime parse and evaluate data in each row. Now all the formatting of data and validations come from the back-end database and we have it in a data-table (dtValidateAndFormatConditions). The conditions would be around 50. So you can see how slow this whole process becomes since N X 150 X 50 operations are required to complete this whole process. Initially it took approximately 2-3 minutes but now i have reduced it to 20 - 30 seconds. However i have increased the speed by making an expression parser of my own - and not by any algorithm, is there any other way i can improve performance, by using Divide and Conquer or some other mechanism. Currently i am not really sure how to go about this. Here is what part of my code looks like: public virtual void ValidateAndFormatOnCopyPaste(DataTable DtCopied, int CurRow) { foreach (DataRow dRow in dtValidateAndFormatConditions.Rows) { string Condition = dRow["Condition"]; string FormatValue = Value = dRow["Value"]; GetValidatedFormattedData(DtCopied,ref Condition, ref FormatValue ,iRowIndex); Condition = Parse(Condition); dRow["Condition"] = Condition; FormatValue = Parse(FormatValue ); dRow["Value"] = FormatValue; } } The above code gets called row-wise like this: public override void ValidateAndFormat(DataTable dtChangedRecords, CellRange cr) { int iRowStart = cr.Row, iRowEnd = cr.Row + cr.RowCount; for (int iRow = iRowStart; iRow < iRowEnd; iRow++) { ValidateAndFormatOnCopyPaste(dtChangedRecords,iRow); } } Please know my question needs a more algorithmic solution than code optimization, however any answers containing code related optimizations will be appreciated as well. (Tagged Linq because although not seen i have been using linq in some parts of my code).

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• Finding the most frequent subtrees in a collection of (parse) trees

  - by peter.murray.rust

  I have a collection of trees whose nodes are labelled (but not uniquely). Specifically the trees are from a collection of parsed sentences (see http://en.wikipedia.org/wiki/Treebank). I wish to extract the most common subtrees from the collection - performance is not (yet) an issue. I'd be grateful for algorithms (ideally Java) or pointers to tools which do this for treebanks. Note that order of child nodes is important. EDIT @mjv. We are working in a limited domain (chemistry) which has a stylised language so the varirty of the trees is not huge - probably similar to children's readers. Simple tree for "the cat sat on the mat". <sentence> <nounPhrase> <article/> <noun/> </nounPhrase> <verbPhrase> <verb/> <prepositionPhrase> <preposition/> <nounPhrase> <article/> <noun/> </nounPhrase> </prepositionPhrase> </verbPhrase> </sentence> Here the sentence contains two identical part-of-speech subtrees (the actual tokens "cat". "mat" are not important in matching). So the algorithm would need to detect this. Note that not all nounPhrases are identical - "the big black cat" could be: <nounPhrase> <article/> <adjective/> <adjective/> <noun/> </nounPhrase> The length of sentences will be longer - between 15 to 30 nodes. I would expect to get useful results from 1000 trees. If this does not take more than a day or so that's acceptable. Obviously the shorter the tree the more frequent, so nounPhrase will be very common. EDIT If this is to be solved by flattening the tree then I think it would be related to Longest Common Substring, not Longest Common Sequence. But note that I don't necessarily just want the longest - I want a list of all those long enough to be "interesting" (criterion yet to be decided).

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• Throwing cats out of windows

  - by AndrewF

  Imagine you're in a tall building with a cat. The cat can survive a fall out of a low story window, but will die if thrown from a high floor. How can you figure out the longest drop that the cat can survive, using the least number of attempts? Obviously, if you only have one cat, then you can only search linearly. First throw the cat from the first floor. If it survives, throw it from the second. Eventually, after being thrown from floor f, the cat will die. You then know that floor f-1 was the maximal safe floor. But what if you have more than one cat? You can now try some sort of logarithmic search. Let's say that the build has 100 floors and you have two identical cats. If you throw the first cat out of the 50th floor and it dies, then you only have to search 50 floors linearly. You can do even better if you choose a lower floor for your first attempt. Let's say that you choose to tackle the problem 20 floors at a time and that the first fatal floor is #50. In that case, your first cat will survive flights from floors 20 and 40 before dying from floor 60. You just have to check floors 41 through 49 individually. That's a total of 12 attempts, which is much better than the 50 you would need had you attempted to use binary elimination. In general, what's the best strategy and it's worst-case complexity for an n-storied building with 2 cats? What about for n floors and m cats? Assume that all cats are equivalent: they will all survive or die from a fall from a given window. Also, every attempt is independent: if a cat survives a fall, it is completely unharmed. This isn't homework, although I may have solved it for school assignment once. It's just a whimsical problem that popped into my head today and I don't remember the solution. Bonus points if anyone knows the name of this problem or of the solution algorithm.

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• Calculate minimum moves to solve a puzzle

  - by Luke

  I'm in the process of creating a game where the user will be presented with 2 sets of colored tiles. In order to ensure that the puzzle is solvable, I start with one set, copy it to a second set, then swap tiles from one set to another. Currently, (and this is where my issue lies) the number of swaps is determined by the level the user is playing - 1 swap for level 1, 2 swaps for level 2, etc. This same number of swaps is used as a goal in the game. The user must complete the puzzle by swapping a tile from one set to the other to make the 2 sets match (by color). The order of the tiles in the (user) solved puzzle doesn't matter as long as the 2 sets match. The problem I have is that as the number of swaps I used to generate the puzzle approaches the number of tiles in each set, the puzzle becomes easier to solve. Basically, you can just drag from one set in whatever order you need for the second set and solve the puzzle with plenty of moves left. What I am looking to do is after I finish building the puzzle, calculate the minimum number of moves required to solve the puzzle. Again, this is almost always less than the number of swaps used to create the puzzle, especially as the number of swaps approaches the number of tiles in each set. My goal is to calculate the best case scenario and then give the user a "fudge factor" (i.e. 1.2 times the minimum number of moves). Solving the puzzle in under this number of moves will result in passing the level. A little background as to how I currently have the game configured: Levels 1 to 10: 9 tiles in each set. 5 different color tiles. Levels 11 to 20: 12 tiles in each set. 7 different color tiles. Levels 21 to 25: 15 tiles in each set. 10 different color tiles. Swapping within a set is not allowed. For each level, there will be at least 2 tiles of a given color (one for each set in the solved puzzle). Is there any type of algorithm anyone could recommend to calculate the minimum number of moves to solve a given puzzle?

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• Calendar Day View in PHP

  - by JamesArmes

  I'm working on adding a day view option to an existing calendar solution. Like many people implementing their own calendars, I am trying to model Google Calendars. They have an excellent calendar solution and their day view provides a lot of flexibility. For most part, the implementation is going well; however, I'm having issues when it comes to conflicting events. Essentially, I want the events to share the same space, side by side. Events that start at the same time should have the longest event first. In the example data set I'm working with, I have four events: A: 10:30 - 11:30 B: 13:30 - 14:30 C: 10:30 - 11:00 D: 10:45 - 14:00 I can handle A, C, and D just fine, the problem comes with D. A should be left of C which should be left of D; each taking one third of the width (it's fixed width so I can do simple math to figure that out). The problem is that B should be under A and C, to the left of D. Ideally, B would take up the same amount of space as A and C (two thirds width), but I would even settle for it only taking up only one third width. My array of events looks similar to the following: $events = array( '1030' => array( 'uniqueID1' => array( 'start_time' => '1030', 'end_time' => '1130', ), 'uniqueID2' => array( 'start_time' => '1030', 'end_time' => '1100', ), ), '1045' => array( 'uniqueID3' => array( 'start_time' => '1045', 'end_time' => '1400', ), ), '1330' => array( 'uniqueID3' => array( 'start_time' => '1330', 'end_time' => '1430', ), ), ); My plan is to add some indexes to each event that include how many events it conflicts with (so I can calculate the width) and which position in that series it should be (so I can calculate the left value). However, that doesn't help the B. I'm thinking I might need an algorithm that uses some basic geometry and matrices, but I'm not sure where to begin. Any help is greatly appreciated.

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• Symfony, in remote host: Error 500. Unknown record property / related component "algorithm" on "sfGu

  - by user248959

  Hi, after deploying, i get the error below after loggingin. Sf 1.3, sfDoctrineGuardPlugin. And i have this schema.yml in config/doctrine: Usuario: inheritance: extends: sfGuardUser type: simple columns: username: type: string(128) notnull: false unique: true nombre_apellidos: string(60) sexo: string(5) fecha_nac: date provincia: string(60) localidad: string(255) email_address: string(255) fotografia: string(255) avatar: string(255) avatar_mensajes: string(255) relations: Usuario: local: user1_id foreign: user2_id refClass: AmigoUsuario equal: true 500 | Internal Server Error | Doctrine_Record_UnknownPropertyException Unknown record property / related component "algorithm" on "sfGuardUser" stack trace * at () in SF_ROOT_DIR/lib/vendor/symfony/lib/plugins/sfDoctrinePlugin/lib/vendor/doctrine/Doctrine/Record/Filter/Standard.php line 55 ... 52. */ 53. public function filterGet(Doctrine_Record $record, $name) 54. { 55. throw new Doctrine_Record_UnknownPropertyException(sprintf('Unknown record property / related component "%s" on "%s"', $name, get_class($record))); 56. } 57. } * at Doctrine_Record_Filter_Standard->filterGet(object('sfGuardUser'), 'algorithm') in SF_ROOT_DIR/lib/vendor/symfony/lib/plugins/sfDoctrinePlugin/lib/vendor/doctrine/Doctrine/Record.php line 1382 ... 1379. $success = false; 1380. foreach ($this->_table->getFilters() as $filter) { 1381. try { 1382. $value = $filter->filterGet($this, $fieldName); 1383. $success = true; 1384. } catch (Doctrine_Exception $e) {} 1385. } * at Doctrine_Record->_get('algorithm', 1) in SF_ROOT_DIR/lib/vendor/symfony/lib/plugins/sfDoctrinePlugin/lib/vendor/doctrine/Doctrine/Record.php line 1337 ... 1334. return $this->$accessor($load); 1335. } 1336. } 1337. return $this->_get($fieldName, $load); 1338. } 1339. 1340. protected function _get($fieldName, $load = true) * at Doctrine_Record->get('algorithm') in SF_ROOT_DIR/lib/vendor/symfony/lib/plugins/sfDoctrinePlugin/lib/record/sfDoctrineRecord.class.php line 212 ... 209. return call_user_func_array( 210. array($this, $verb), 211. array_merge(array($entityName), $arguments) 212. ); 213. } else { 214. $failed = true; 215. } * at sfDoctrineRecord->__call(array(object('sfGuardUser'), 'get'), array('algorithm')) in n/a line n/a ... * at sfGuardUser->getAlgorithm('getAlgorithm', array()) in SF_ROOT_DIR/plugins/sfDoctrineGuardPlugin/lib/model/doctrine/PluginsfGuardUser.class.php line 96 ... 93. */ 94. public function checkPasswordByGuard($password) 95. { 96. $algorithm = $this->getAlgorithm(); 97. if (false !== $pos = strpos($algorithm, '::')) 98. { 99. $algorithm = array(substr($algorithm, 0, $pos), substr($algorithm, $pos + 2)); * at PluginsfGuardUser->checkPasswordByGuard() in SF_ROOT_DIR/plugins/sfDoctrineGuardPlugin/lib/model/doctrine/PluginsfGuardUser.class.php line 83 ... 80. } 81. else 82. { 83. return $this->checkPasswordByGuard($password); 84. } 85. } 86. * at PluginsfGuardUser->checkPassword('m') in SF_ROOT_DIR/lib/sfGuardValidatorUserByEmail.class.php line 28 ... 25. { 26. // password is ok? 27. 28. if ($user->checkPassword($password)) 29. { 30. 31. //die("entro"); * at sfGuardValidatorUserByEmail->doClean('m') Any idea? Javi

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• Is it possible to shuffle a 2D matrix while preserving row AND column frequencies?

  - by j_random_hacker

  Suppose I have a 2D array like the following: GACTG AGATA TCCGA Each array element is taken from a small finite set (in my case, DNA nucleotides -- {A, C, G, T}). I would like to randomly shuffle this array somehow while preserving both row and column nucleotide frequencies. Is this possible? Can it be done efficiently? [EDIT]: By this I mean I want to produce a new matrix where each row has the same number of As, Cs, Gs and Ts as the corresponding row of the original matrix, and where each column has the same number of As, Cs, Gs and Ts as the corresponding column of the original matrix. Permuting the rows or columns of the original matrix will not achieve this in general. (E.g. for the example above, the top row has 2 Gs, and 1 each of A, C and T; if this row was swapped with row 2, the top row in the resulting matrix would have 3 As, 1 G and 1 T.) It's simple enough to preserve just column frequencies by shuffling a column at a time, and likewise for rows. But doing this will in general alter the frequencies of the other kind. My thoughts so far: If it's possible to pick 2 rows and 2 columns so that the 4 elements at the corners of this rectangle have the pattern XY YX for some pair of distinct elements X and Y, then replacing these 4 elements with YX XY will maintain both row and column frequencies. In the example at the top, this can be done for (at least) rows 1 and 2 and columns 2 and 5 (whose corners give the 2x2 matrix AG;GA), and for rows 1 and 3 and columns 1 and 4 (whose corners give GT;TG). Clearly this could be repeated a number of times to produce some level of randomisation. Generalising a bit, any "subrectangle" induced by a subset of rows and a subset of columns, in which the frequencies of all rows are the same and the frequencies of all columns are the same, can have both its rows and columns permuted to produce a valid complete rectangle. (Of these, only those subrectangles in which at least 1 element is changed are actually interesting.) Big questions: Are all valid complete matrices reachable by a series of such "subrectangle rearrangements"? I suspect the answer is yes. Are all valid subrectangle rearrangements decomposable into a series of 2x2 swaps? I suspect the answer is no, but I hope it's yes, since that would seem to make it easier to come up with an efficient algorithm. Can some or all of the valid rearrangements be computed efficiently? This question addresses a special case in which the set of possible elements is {0, 1}. The solutions people have come up with there are similar to what I have come up with myself, and are probably usable, but not ideal as they require an arbitrary amount of backtracking to work correctly. Also I'm concerned that only 2x2 swaps are considered. Finally, I would ideally like a solution that can be proven to select a matrix uniformly at random from the set of all matrices having identical row frequencies and column frequencies to the original. I know, I'm asking for a lot :)

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• Calculate the number of ways to roll a certain number

  - by helloworld

  I'm a high school Computer Science student, and today I was given a problem to: Program Description: There is a belief among dice players that in throwing three dice a ten is easier to get than a nine. Can you write a program that proves or disproves this belief? Have the computer compute all the possible ways three dice can be thrown: 1 + 1 + 1, 1 + 1 + 2, 1 + 1 + 3, etc. Add up each of these possibilities and see how many give nine as the result and how many give ten. If more give ten, then the belief is proven. I quickly worked out a brute force solution, as such int sum,tens,nines; tens=nines=0; for(int i=1;i<=6;i++){ for(int j=1;j<=6;j++){ for(int k=1;k<=6;k++){ sum=i+j+k; //Ternary operators are fun! tens+=((sum==10)?1:0); nines+=((sum==9)?1:0); } } } System.out.println("There are "+tens+" ways to roll a 10"); System.out.println("There are "+nines+" ways to roll a 9"); Which works just fine, and a brute force solution is what the teacher wanted us to do. However, it doesn't scale, and I am trying to find a way to make an algorithm that can calculate the number of ways to roll n dice to get a specific number. Therefore, I started generating the number of ways to get each sum with n dice. With 1 die, there is obviously 1 solution for each. I then calculated, through brute force, the combinations with 2 and 3 dice. These are for two: There are 1 ways to roll a 2 There are 2 ways to roll a 3 There are 3 ways to roll a 4 There are 4 ways to roll a 5 There are 5 ways to roll a 6 There are 6 ways to roll a 7 There are 5 ways to roll a 8 There are 4 ways to roll a 9 There are 3 ways to roll a 10 There are 2 ways to roll a 11 There are 1 ways to roll a 12 Which looks straightforward enough; it can be calculated with a simple linear absolute value function. But then things start getting trickier. With 3: There are 1 ways to roll a 3 There are 3 ways to roll a 4 There are 6 ways to roll a 5 There are 10 ways to roll a 6 There are 15 ways to roll a 7 There are 21 ways to roll a 8 There are 25 ways to roll a 9 There are 27 ways to roll a 10 There are 27 ways to roll a 11 There are 25 ways to roll a 12 There are 21 ways to roll a 13 There are 15 ways to roll a 14 There are 10 ways to roll a 15 There are 6 ways to roll a 16 There are 3 ways to roll a 17 There are 1 ways to roll a 18 So I look at that, and I think: Cool, Triangular numbers! However, then I notice those pesky 25s and 27s. So it's obviously not triangular numbers, but still some polynomial expansion, since it's symmetric. So I take to Google, and I come across this page that goes into some detail about how to do this with math. It is fairly easy(albeit long) to find this using repeated derivatives or expansion, but it would be much harder to program that for me. I didn't quite understand the second and third answers, since I have never encountered that notation or those concepts in my math studies before. Could someone please explain how I could write a program to do this, or explain the solutions given on that page, for my own understanding of combinatorics? EDIT: I'm looking for a mathematical way to solve this, that gives an exact theoretical number, not by simulating dice

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• Testing for Adjacent Cells In a Multi-level Grid

  - by Steve

  I'm designing an algorithm to test whether cells on a grid are adjacent or not. The catch is that the cells are not on a flat grid. They are on a multi-level grid such as the one drawn below. Level 1 (Top Level) | - - - - - | | A | B | C | | - - - - - | | D | E | F | | - - - - - | | G | H | I | | - - - - - | Level 2 | -Block A- | -Block B- | | 1 | 2 | 3 | 1 | 2 | 3 | | - - - - - | - - - - - | | 4 | 5 | 6 | 4 | 5 | 6 | ... | - - - - - | - - - - - | | 7 | 8 | 9 | 7 | 8 | 9 | | - - - - - | - - - - - | | -Block D- | -Block E- | | 1 | 2 | 3 | 1 | 2 | 3 | | - - - - - | - - - - - | | 4 | 5 | 6 | 4 | 5 | 6 | ... | - - - - - | - - - - - | | 7 | 8 | 9 | 7 | 8 | 9 | | - - - - - | - - - - - | . . . . . . This diagram is simplified from my actual need but the concept is the same. There is a top level block with many cells within it (level 1). Each block is further subdivided into many more cells (level 2). Those cells are further subdivided into level 3, 4 and 5 for my project but let's just stick to two levels for this question. I'm receiving inputs for my function in the form of "A8, A9, B7, D3". That's a list of cell Ids where each cell Id has the format (level 1 id)(level 2 id). Let's start by comparing just 2 cells, A8 and A9. That's easy because they are in the same block. private static RelativePosition getRelativePositionInTheSameBlock(String v1, String v2) { RelativePosition relativePosition; if( v1-v2 == -1 ) { relativePosition = RelativePosition.LEFT_OF; } else if (v1-v2 == 1) { relativePosition = RelativePosition.RIGHT_OF; } else if (v1-v2 == -BLOCK_WIDTH) { relativePosition = RelativePosition.TOP_OF; } else if (v1-v2 == BLOCK_WIDTH) { relativePosition = RelativePosition.BOTTOM_OF; } else { relativePosition = RelativePosition.NOT_ADJACENT; } return relativePosition; } An A9 - B7 comparison could be done by checking if A is a multiple of BLOCK_WIDTH and whether B is (A-BLOCK_WIDTH+1). Either that or just check naively if the A/B pair is 3-1, 6-4 or 9-7 for better readability. For B7 - D3, they are not adjacent but D3 is adjacent to A9 so I can do a similar adjacency test as above. So getting away from the little details and focusing on the big picture. Is this really the best way to do it? Keeping in mind the following points: I actually have 5 levels not 2, so I could potentially get a list like "A8A1A, A8A1B, B1A2A, B1A2B". Adding a new cell to compare still requires me to compare all the other cells before it (seems like the best I could do for this step is O(n)) The cells aren't all 3x3 blocks, they're just that way for my example. They could be MxN blocks with different M and N for different levels. In my current implementation above, I have separate functions to check adjacency if the cells are in the same blocks, if they are in separate horizontally adjacent blocks or if they are in separate vertically adjacent blocks. That means I have to know the position of the two blocks at the current level before I call one of those functions for the layer below. Judging by the complexity of having to deal with mulitple functions for different edge cases at different levels and having 5 levels of nested if statements. I'm wondering if another design is more suitable. Perhaps a more recursive solution, use of other data structures, or perhaps map the entire multi-level grid to a single-level grid (my quick calculations gives me about 700,000+ atomic cell ids). Even if I go that route, mapping from multi-level to single level is a non-trivial task in itself.

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• Prim's MST algorithm implementation with Java

  - by user1290164

  I'm trying to write a program that'll find the MST of a given undirected weighted graph with Kruskal's and Prim's algorithms. I've successfully implemented Kruskal's algorithm in the program, but I'm having trouble with Prim's. To be more precise, I can't figure out how to actually build the Prim function so that it'll iterate through all the vertices in the graph. I'm getting some IndexOutOfBoundsException errors during program execution. I'm not sure how much information is needed for others to get the idea of what I have done so far, but hopefully there won't be too much useless information. This is what I have so far: I have a Graph, Edge and a Vertex class. Vertex class mostly just an information storage that contains the name (number) of the vertex. Edge class can create a new Edge that has gets parameters (Vertex start, Vertex end, int edgeWeight). The class has methods to return the usual info like start vertex, end vertex and the weight. Graph class reads data from a text file and adds new Edges to an ArrayList. The text file also tells us how many vertecis the graph has, and that gets stored too. In the Graph class, I have a Prim() -method that's supposed to calculate the MST: public ArrayList<Edge> Prim(Graph G) { ArrayList<Edge> edges = G.graph; // Copies the ArrayList with all edges in it. ArrayList<Edge> MST = new ArrayList<Edge>(); Random rnd = new Random(); Vertex startingVertex = edges.get(rnd.nextInt(G.returnVertexCount())).returnStartingVertex(); // This is just to randomize the starting vertex. // This is supposed to be the main loop to find the MST, but this is probably horribly wrong.. while (MST.size() < returnVertexCount()) { Edge e = findClosestNeighbour(startingVertex); MST.add(e); visited.add(e.returnStartingVertex()); visited.add(e.returnEndingVertex()); edges.remove(e); } return MST; } The method findClosesNeighbour() looks like this: public Edge findClosestNeighbour(Vertex v) { ArrayList<Edge> neighbours = new ArrayList<Edge>(); ArrayList<Edge> edges = graph; for (int i = 0; i < edges.size() -1; ++i) { if (edges.get(i).endPoint() == s.returnVertexID() && !visited(edges.get(i).returnEndingVertex())) { neighbours.add(edges.get(i)); } } return neighbours.get(0); // This is the minimum weight edge in the list. } ArrayList<Vertex> visited and ArrayList<Edges> graph get constructed when creating a new graph. Visited() -method is simply a boolean check to see if ArrayList visited contains the Vertex we're thinking about moving to. I tested the findClosestNeighbour() independantly and it seemed to be working but if someone finds something wrong with it then that feedback is welcome also. Mainly though as I mentioned my problem is with actually building the main loop in the Prim() -method, and if there's any additional info needed I'm happy to provide it. Thank you. Edit: To clarify what my train of thought with the Prim() method is. What I want to do is first randomize the starting point in the graph. After that, I will find the closest neighbor to that starting point. Then we'll add the edge connecting those two points to the MST, and also add the vertices to the visited list for checking later, so that we won't form any loops in the graph. Here's the error that gets thrown: Exception in thread "main" java.lang.IndexOutOfBoundsException: Index: 0, Size: 0 at java.util.ArrayList.rangeCheck(Unknown Source) at java.util.ArrayList.get(Unknown Source) at Graph.findClosestNeighbour(graph.java:203) at Graph.Prim(graph.java:179) at MST.main(MST.java:49) Line 203: return neighbour.get(0); in findClosestNeighbour() Line 179: Edge e = findClosestNeighbour(startingVertex); in Prim()

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• Gradient algororithm produces little white dots

  - by user146780

  I'm working on an algorithm to generate point to point linear gradients. I have a rough, proof of concept implementation done: GLuint OGLENGINEFUNCTIONS::CreateGradient( std::vector<ARGBCOLORF> &input,POINTFLOAT start, POINTFLOAT end, int width, int height,bool radial ) { std::vector<POINT> pol; std::vector<GLubyte> pdata(width * height * 4); std::vector<POINTFLOAT> linearpts; std::vector<float> lookup; float distance = GetDistance(start,end); RoundNumber(distance); POINTFLOAT temp; float incr = 1 / (distance + 1); for(int l = 0; l < 100; l ++) { POINTFLOAT outA; POINTFLOAT OutB; float dirlen; float perplen; POINTFLOAT dir; POINTFLOAT ndir; POINTFLOAT perp; POINTFLOAT nperp; POINTFLOAT perpoffset; POINTFLOAT diroffset; dir.x = end.x - start.x; dir.y = end.y - start.y; dirlen = sqrt((dir.x * dir.x) + (dir.y * dir.y)); ndir.x = static_cast<float>(dir.x * 1.0 / dirlen); ndir.y = static_cast<float>(dir.y * 1.0 / dirlen); perp.x = dir.y; perp.y = -dir.x; perplen = sqrt((perp.x * perp.x) + (perp.y * perp.y)); nperp.x = static_cast<float>(perp.x * 1.0 / perplen); nperp.y = static_cast<float>(perp.y * 1.0 / perplen); perpoffset.x = static_cast<float>(nperp.x * l * 0.5); perpoffset.y = static_cast<float>(nperp.y * l * 0.5); diroffset.x = static_cast<float>(ndir.x * 0 * 0.5); diroffset.y = static_cast<float>(ndir.y * 0 * 0.5); outA.x = end.x + perpoffset.x + diroffset.x; outA.y = end.y + perpoffset.y + diroffset.y; OutB.x = start.x + perpoffset.x - diroffset.x; OutB.y = start.y + perpoffset.y - diroffset.y; for (float i = 0; i < 1; i += incr) { temp = GetLinearBezier(i,outA,OutB); RoundNumber(temp.x); RoundNumber(temp.y); linearpts.push_back(temp); lookup.push_back(i); } for (unsigned int j = 0; j < linearpts.size(); j++) { if(linearpts[j].x < width && linearpts[j].x >= 0 && linearpts[j].y < height && linearpts[j].y >=0) { pdata[linearpts[j].x * 4 * width + linearpts[j].y * 4 + 0] = (GLubyte) j; pdata[linearpts[j].x * 4 * width + linearpts[j].y * 4 + 1] = (GLubyte) j; pdata[linearpts[j].x * 4 * width + linearpts[j].y * 4 + 2] = (GLubyte) j; pdata[linearpts[j].x * 4 * width + linearpts[j].y * 4 + 3] = (GLubyte) 255; } } lookup.clear(); linearpts.clear(); } return CreateTexture(pdata,width,height); } It works as I would expect most of the time, but at certain angles it produces little white dots. I can't figure out what does this. This is what it looks like at most angles (good) http://img9.imageshack.us/img9/5922/goodgradient.png But once in a while it looks like this (bad): http://img155.imageshack.us/img155/760/badgradient.png What could be causing the white dots? Is there maybe also a better way to generate my gradients if no solution is possible for this? Thanks

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• What is wrong with my logic for the divide and conquer algorithm for Closest pair problem?

  - by Programming Noob

  I have been following Coursera's course on Algorithms and came up with a thought about the divide/conquer algorithm for the closest pair problem, that I want clarified. As per Prof Roughgarden's algorithm (which you can see here if you're interested): For a given set of points P, of which we have two copies - sorted in X and Y direction - Px and Py, the algorithm can be given as closestPair(Px,Py): Divide points into left half - Q, and right half - R, and form sorted copies of both halves along x and y directions - Qx,Qy,Rx,Ry Let closestPair(Qx,Qy) be points p1 and q1 Let closestPair(Rx,Ry) be p2,q2 Let delta be minimum of dist(p1,q1) and dist(p2,q2) This is the unfortunate case, let p3,q3 be the closestSplitPair(Px,Py,delta) Return the best result Now, the clarification that I want is related to step 5. I should say this beforehand, that what I'm suggesting, is barely any improvement at all, but if you're still interested, read ahead. Prof R says that since the points are already sorted in X and Y directions, to find the best pair in step 5, we need to iterate over points in the strip of width 2*delta, starting from bottom to up, and in the inner loop we need only 7 comparisions. Can this be bettered to just one? How I think is possible seemed a little difficult to explain in plain text, so I drew a diagram and wrote it on paper and uploaded it here: Since no one else came up with is, I'm pretty sure there's some error in my line of thought. But I have literally been thinking about this for HOURS now, and I just HAD to post this. It's all that is in my head. Can someone point out where I'm going wrong?

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• If your algorithm is correct, does it matter how long it took you to write it?

  - by John Isaacks

  I recently found out that Facebook had a programming challenge that if completed correctly you automatically get a phone interview. There is a sample challenge that asks you to write an algorithm that can solve a Tower of Hanoi type problem. Given a number of pegs and discs, an initial and final configuration; Your algorithm must determine the fewest steps possible to get to the final configuration and output the steps. This sample challenge gives you a 45 minute time limit but allows you to still test your code to see if it passes once your time limit expires. I did not know of any cute math solution that could solve it, and I didn't want to look for one since I think that would be cheating. So I tried to solve the challenge the best I could on my own. I was able to make an algorithm that worked and passed. However, it took me over 4 hours to make, much longer than the 45 minute requirement. Since it took me so much longer than the allotted time, I have not attempted the actual challenge. This got me wondering though, in reality does it really matter that it took me that long? I mean is this a sign that I will not be able to get a job at a place like this (not just Facebook, but Google, Fog Creek, etc.) and need to lower my aspirations, or does the fact that I actually passed on my first attempt even though it took too long be taken as good?

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• questions regarding the use of A* with the 15-square puzzle

  - by Cheeso

  I'm trying to build an A* solver for a 15-square puzzle. The goal is to re-arrange the tiles so that they appear in their natural positions. You can only slide one tile at a time. Each possible state of the puzzle is a node in the search graph. For the h(x) function, I am using an aggregate sum, across all tiles, of the tile's dislocation from the goal state. In the above image, the 5 is at location 0,0, and it belongs at location 1,0, therefore it contributes 1 to the h(x) function. The next tile is the 11, located at 0,1, and belongs at 2,2, therefore it contributes 3 to h(x). And so on. EDIT: I now understand this is what they call "Manhattan distance", or "taxicab distance". I have been using a step count for g(x). In my implementation, for any node in the state graph, g is just +1 from the prior node's g. To find successive nodes, I just examine where I can possibly move the "hole" in the puzzle. There are 3 neighbors for the puzzle state (aka node) that is displayed: the hole can move north, west, or east. My A* search sometimes converges to a solution in 20s, sometimes 180s, and sometimes doesn't converge at all (waited 10 mins or more). I think h is reasonable. I'm wondering if I've modeled g properly. In other words, is it possible that my A* function is reaching a node in the graph via a path that is not the shortest path? Maybe have I not waited long enough? Maybe 10 minutes is not long enough? For a fully random arrangement, (assuming no parity problems), What is the average number of permutations an A* solution will examine? (please show the math) I'm going to look for logic errors in my code, but in the meantime, Any tips? (ps: it's done in Javascript). Also, no, this isn't CompSci homework. It's just a personal exploration thing. I'm just trying to learn Javascript. EDIT: I've found that the run-time is highly depend upon the heuristic. I saw the 10x factor applied to the heuristic from the article someone mentioned, and it made me wonder - why 10x? Why linear? Because this is done in javascript, I could modify the code to dynamically update an html table with the node currently being considered. This allowd me to peek at the algorithm as it was progressing. With a regular taxicab distance heuristic, I watched as it failed to converge. There were 5's and 12's in the top row, and they kept hanging around. I'd see 1,2,3,4 creep into the top row, but then they'd drop out, and other numbers would move up there. What I was hoping to see was 1,2,3,4 sort of creeping up to the top, and then staying there. I thought to myself - this is not the way I solve this personally. Doing this manually, I solve the top row, then the 2ne row, then the 3rd and 4th rows sort of concurrently. So I tweaked the h(x) function to more heavily weight the higher rows and the "lefter" columns. The result was that the A* converged much more quickly. It now runs in 3 minutes instead of "indefinitely". With the "peek" I talked about, I can see the smaller numbers creep up to the higher rows and stay there. Not only does this seem like the right thing, it runs much faster. I'm in the process of trying a bunch of variations. It seems pretty clear that A* runtime is very sensitive to the heuristic. Currently the best heuristic I've found uses the summation of dislocation * ((4-i) + (4-j)) where i and j are the row and column, and dislocation is the taxicab distance. One interesting part of the result I got: with a particular heuristic I find a path very quickly, but it is obviously not the shortest path. I think this is because I am weighting the heuristic. In one case I got a path of 178 steps in 10s. My own manual effort produce a solution in 87 moves. (much more than 10s). More investigation warranted. So the result is I am seeing it converge must faster, and the path is definitely not the shortest. I have to think about this more. Code: var stop = false; function Astar(start, goal, callback) { // start and goal are nodes in the graph, represented by // an array of 16 ints. The goal is: [1,2,3,...14,15,0] // Zero represents the hole. // callback is a method to call when finished. This runs a long time, // therefore we need to use setTimeout() to break it up, to avoid // the browser warning like "Stop running this script?" // g is the actual distance traveled from initial node to current node. // h is the heuristic estimate of distance from current to goal. stop = false; start.g = start.dontgo = 0; // calcHeuristic inserts an .h member into the array calcHeuristicDistance(start); // start the stack with one element var closed = []; // set of nodes already evaluated. var open = [ start ]; // set of nodes to evaluate (start with initial node) var iteration = function() { if (open.length==0) { // no more nodes. Fail. callback(null); return; } var current = open.shift(); // get highest priority node // update the browser with a table representation of the // node being evaluated $("#solution").html(stateToString(current)); // check solution returns true if current == goal if (checkSolution(current,goal)) { // reconstructPath just records the position of the hole // through each node var path= reconstructPath(start,current); callback(path); return; } closed.push(current); // get the set of neighbors. This is 3 or fewer nodes. // (nextStates is optimized to NOT turn directly back on itself) var neighbors = nextStates(current, goal); for (var i=0; i<neighbors.length; i++) { var n = neighbors[i]; // skip this one if we've already visited it if (closed.containsNode(n)) continue; // .g, .h, and .previous get assigned implicitly when // calculating neighbors. n.g is nothing more than // current.g+1 ; // add to the open list if (!open.containsNode(n)) { // slot into the list, in priority order (minimum f first) open.priorityPush(n); n.previous = current; } } if (stop) { callback(null); return; } setTimeout(iteration, 1); }; // kick off the first iteration iteration(); return null; }

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• Need help with fixing Genetic Algorithm that's not evolving correctly

  - by EnderMB

  I am working on a maze solving application that uses a Genetic Algorithm to evolve a set of genes (within Individuals) to evolve a Population of Individuals that power an Agent through a maze. The majority of the code used appears to be working fine but when the code runs it's not selecting the best Individual's to be in the new Population correctly. When I run the application it outputs the following: Total Fitness: 380.0 - Best Fitness: 11.0 Total Fitness: 406.0 - Best Fitness: 15.0 Total Fitness: 344.0 - Best Fitness: 12.0 Total Fitness: 373.0 - Best Fitness: 11.0 Total Fitness: 415.0 - Best Fitness: 12.0 Total Fitness: 359.0 - Best Fitness: 11.0 Total Fitness: 436.0 - Best Fitness: 13.0 Total Fitness: 390.0 - Best Fitness: 12.0 Total Fitness: 379.0 - Best Fitness: 15.0 Total Fitness: 370.0 - Best Fitness: 11.0 Total Fitness: 361.0 - Best Fitness: 11.0 Total Fitness: 413.0 - Best Fitness: 16.0 As you can clearly see the fitnesses are not improving and neither are the best fitnesses. The main code responsible for this problem is here, and I believe the problem to be within the main method, most likely where the selection methods are called: package GeneticAlgorithm; import GeneticAlgorithm.Individual.Action; import Robot.Robot.Direction; import Maze.Maze; import Robot.Robot; import java.util.ArrayList; import java.util.Random; public class RunGA { protected static ArrayList tmp1, tmp2 = new ArrayList(); // Implementation of Elitism protected static int ELITISM_K = 5; // Population size protected static int POPULATION_SIZE = 50 + ELITISM_K; // Max number of Iterations protected static int MAX_ITERATIONS = 200; // Probability of Mutation protected static double MUTATION_PROB = 0.05; // Probability of Crossover protected static double CROSSOVER_PROB = 0.7; // Instantiate Random object private static Random rand = new Random(); // Instantiate Population of Individuals private Individual[] startPopulation; // Total Fitness of Population private double totalFitness; Robot robot = new Robot(); Maze maze; public void setElitism(int result) { ELITISM_K = result; } public void setPopSize(int result) { POPULATION_SIZE = result + ELITISM_K; } public void setMaxIt(int result) { MAX_ITERATIONS = result; } public void setMutProb(double result) { MUTATION_PROB = result; } public void setCrossoverProb(double result) { CROSSOVER_PROB = result; } /** * Constructor for Population */ public RunGA(Maze maze) { // Create a population of population plus elitism startPopulation = new Individual[POPULATION_SIZE]; // For every individual in population fill with x genes from 0 to 1 for (int i = 0; i < POPULATION_SIZE; i++) { startPopulation[i] = new Individual(); startPopulation[i].randGenes(); } // Evaluate the current population's fitness this.evaluate(maze, startPopulation); } /** * Set Population * @param newPop */ public void setPopulation(Individual[] newPop) { System.arraycopy(newPop, 0, this.startPopulation, 0, POPULATION_SIZE); } /** * Get Population * @return */ public Individual[] getPopulation() { return this.startPopulation; } /** * Evaluate fitness * @return */ public double evaluate(Maze maze, Individual[] newPop) { this.totalFitness = 0.0; ArrayList<Double> fitnesses = new ArrayList<Double>(); for (int i = 0; i < POPULATION_SIZE; i++) { maze = new Maze(8, 8); maze.fillMaze(); fitnesses.add(startPopulation[i].evaluate(maze, newPop)); //this.totalFitness += startPopulation[i].evaluate(maze, newPop); } //totalFitness = (Math.round(totalFitness / POPULATION_SIZE)); StringBuilder sb = new StringBuilder(); for(Double tmp : fitnesses) { sb.append(tmp + ", "); totalFitness += tmp; } // Progress of each Individual //System.out.println(sb.toString()); return this.totalFitness; } /** * Roulette Wheel Selection * @return */ public Individual rouletteWheelSelection() { // Calculate sum of all chromosome fitnesses in population - sum S. double randNum = rand.nextDouble() * this.totalFitness; int i; for (i = 0; i < POPULATION_SIZE && randNum > 0; ++i) { randNum -= startPopulation[i].getFitnessValue(); } return startPopulation[i-1]; } /** * Tournament Selection * @return */ public Individual tournamentSelection() { double randNum = rand.nextDouble() * this.totalFitness; // Get random number of population (add 1 to stop nullpointerexception) int k = rand.nextInt(POPULATION_SIZE) + 1; int i; for (i = 1; i < POPULATION_SIZE && i < k && randNum > 0; ++i) { randNum -= startPopulation[i].getFitnessValue(); } return startPopulation[i-1]; } /** * Finds the best individual * @return */ public Individual findBestIndividual() { int idxMax = 0; double currentMax = 0.0; double currentMin = 1.0; double currentVal; for (int idx = 0; idx < POPULATION_SIZE; ++idx) { currentVal = startPopulation[idx].getFitnessValue(); if (currentMax < currentMin) { currentMax = currentMin = currentVal; idxMax = idx; } if (currentVal > currentMax) { currentMax = currentVal; idxMax = idx; } } // Double check to see if this has the right one //System.out.println(startPopulation[idxMax].getFitnessValue()); // Maximisation return startPopulation[idxMax]; } /** * One Point Crossover * @param firstPerson * @param secondPerson * @return */ public static Individual[] onePointCrossover(Individual firstPerson, Individual secondPerson) { Individual[] newPerson = new Individual[2]; newPerson[0] = new Individual(); newPerson[1] = new Individual(); int size = Individual.SIZE; int randPoint = rand.nextInt(size); int i; for (i = 0; i < randPoint; ++i) { newPerson[0].setGene(i, firstPerson.getGene(i)); newPerson[1].setGene(i, secondPerson.getGene(i)); } for (; i < Individual.SIZE; ++i) { newPerson[0].setGene(i, secondPerson.getGene(i)); newPerson[1].setGene(i, firstPerson.getGene(i)); } return newPerson; } /** * Uniform Crossover * @param firstPerson * @param secondPerson * @return */ public static Individual[] uniformCrossover(Individual firstPerson, Individual secondPerson) { Individual[] newPerson = new Individual[2]; newPerson[0] = new Individual(); newPerson[1] = new Individual(); for(int i = 0; i < Individual.SIZE; ++i) { double r = rand.nextDouble(); if (r > 0.5) { newPerson[0].setGene(i, firstPerson.getGene(i)); newPerson[1].setGene(i, secondPerson.getGene(i)); } else { newPerson[0].setGene(i, secondPerson.getGene(i)); newPerson[1].setGene(i, firstPerson.getGene(i)); } } return newPerson; } public double getTotalFitness() { return totalFitness; } public static void main(String[] args) { // Initialise Environment Maze maze = new Maze(8, 8); maze.fillMaze(); // Instantiate Population //Population pop = new Population(); RunGA pop = new RunGA(maze); // Instantiate Individuals for Population Individual[] newPop = new Individual[POPULATION_SIZE]; // Instantiate two individuals to use for selection Individual[] people = new Individual[2]; Action action = null; Direction direction = null; String result = ""; /*result += "Total Fitness: " + pop.getTotalFitness() + " - Best Fitness: " + pop.findBestIndividual().getFitnessValue();*/ // Print Current Population System.out.println("Total Fitness: " + pop.getTotalFitness() + " - Best Fitness: " + pop.findBestIndividual().getFitnessValue()); // Instantiate counter for selection int count; for (int i = 0; i < MAX_ITERATIONS; i++) { count = 0; // Elitism for (int j = 0; j < ELITISM_K; ++j) { // This one has the best fitness newPop[count] = pop.findBestIndividual(); count++; } // Build New Population (Population size = Steps (28)) while (count < POPULATION_SIZE) { // Roulette Wheel Selection people[0] = pop.rouletteWheelSelection(); people[1] = pop.rouletteWheelSelection(); // Tournament Selection //people[0] = pop.tournamentSelection(); //people[1] = pop.tournamentSelection(); // Crossover if (rand.nextDouble() < CROSSOVER_PROB) { // One Point Crossover //people = onePointCrossover(people[0], people[1]); // Uniform Crossover people = uniformCrossover(people[0], people[1]); } // Mutation if (rand.nextDouble() < MUTATION_PROB) { people[0].mutate(); } if (rand.nextDouble() < MUTATION_PROB) { people[1].mutate(); } // Add to New Population newPop[count] = people[0]; newPop[count+1] = people[1]; count += 2; } // Make new population the current population pop.setPopulation(newPop); // Re-evaluate the current population //pop.evaluate(); pop.evaluate(maze, newPop); // Print results to screen System.out.println("Total Fitness: " + pop.totalFitness + " - Best Fitness: " + pop.findBestIndividual().getFitnessValue()); //result += "\nTotal Fitness: " + pop.totalFitness + " - Best Fitness: " + pop.findBestIndividual().getFitnessValue(); } // Best Individual Individual bestIndiv = pop.findBestIndividual(); //return result; } } I have uploaded the full project to RapidShare if you require the extra files, although if needed I can add the code to them here. This problem has been depressing me for days now and if you guys can help me I will forever be in your debt.

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• Invalid algorithm specified on Windows 2003 Server only

  - by JL

  I am decoding a file using the following method: string outFileName = zfoFileName.Replace(".zfo", "_tmp.zfo"); FileStream inFile = null; FileStream outFile = null; inFile = File.Open(zfoFileName, FileMode.Open); outFile = File.Create(outFileName); LargeCMS.CMS cms = new LargeCMS.CMS(); cms.Decode(inFile, outFile); This is working fine on my Win 7 dev machine, but on a Windows 2003 server production machine it fails with the following exception: Exception: System.Exception: CryptMsgUpdate error #-2146893816 --- System.ComponentModel.Win32Exception: Invalid algorithm specified --- End of inner exception stack trace --- at LargeCMS.CMS.Decode(FileStream inFile, FileStream outFile) Here are the classes below which I call to do the decoding, if needed I can upload a sample file for decoding, its just strange it works on Win 7, and not on Win2k3 server: using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.IO; using System.Security.Cryptography; using System.Security.Cryptography.X509Certificates; using System.Runtime.InteropServices; using System.ComponentModel; namespace LargeCMS { class CMS { // File stream to use in callback function private FileStream m_callbackFile; // Streaming callback function for encoding private Boolean StreamOutputCallback(IntPtr pvArg, IntPtr pbData, int cbData, Boolean fFinal) { // Write all bytes to encoded file Byte[] bytes = new Byte[cbData]; Marshal.Copy(pbData, bytes, 0, cbData); m_callbackFile.Write(bytes, 0, cbData); if (fFinal) { // This is the last piece. Close the file m_callbackFile.Flush(); m_callbackFile.Close(); m_callbackFile = null; } return true; } // Encode CMS with streaming to support large data public void Encode(X509Certificate2 cert, FileStream inFile, FileStream outFile) { // Variables Win32.CMSG_SIGNER_ENCODE_INFO SignerInfo; Win32.CMSG_SIGNED_ENCODE_INFO SignedInfo; Win32.CMSG_STREAM_INFO StreamInfo; Win32.CERT_CONTEXT[] CertContexts = null; Win32.BLOB[] CertBlobs; X509Chain chain = null; X509ChainElement[] chainElements = null; X509Certificate2[] certs = null; RSACryptoServiceProvider key = null; BinaryReader stream = null; GCHandle gchandle = new GCHandle(); IntPtr hProv = IntPtr.Zero; IntPtr SignerInfoPtr = IntPtr.Zero; IntPtr CertBlobsPtr = IntPtr.Zero; IntPtr hMsg = IntPtr.Zero; IntPtr pbPtr = IntPtr.Zero; Byte[] pbData; int dwFileSize; int dwRemaining; int dwSize; Boolean bResult = false; try { // Get data to encode dwFileSize = (int)inFile.Length; stream = new BinaryReader(inFile); pbData = stream.ReadBytes(dwFileSize); // Prepare stream for encoded info m_callbackFile = outFile; // Get cert chain chain = new X509Chain(); chain.Build(cert); chainElements = new X509ChainElement[chain.ChainElements.Count]; chain.ChainElements.CopyTo(chainElements, 0); // Get certs in chain certs = new X509Certificate2[chainElements.Length]; for (int i = 0; i < chainElements.Length; i++) { certs[i] = chainElements[i].Certificate; } // Get context of all certs in chain CertContexts = new Win32.CERT_CONTEXT[certs.Length]; for (int i = 0; i < certs.Length; i++) { CertContexts[i] = (Win32.CERT_CONTEXT)Marshal.PtrToStructure(certs[i].Handle, typeof(Win32.CERT_CONTEXT)); } // Get cert blob of all certs CertBlobs = new Win32.BLOB[CertContexts.Length]; for (int i = 0; i < CertContexts.Length; i++) { CertBlobs[i].cbData = CertContexts[i].cbCertEncoded; CertBlobs[i].pbData = CertContexts[i].pbCertEncoded; } // Get CSP of client certificate key = (RSACryptoServiceProvider)certs[0].PrivateKey; bResult = Win32.CryptAcquireContext( ref hProv, key.CspKeyContainerInfo.KeyContainerName, key.CspKeyContainerInfo.ProviderName, key.CspKeyContainerInfo.ProviderType, 0 ); if (!bResult) { throw new Exception("CryptAcquireContext error #" + Marshal.GetLastWin32Error().ToString(), new Win32Exception(Marshal.GetLastWin32Error())); } // Populate Signer Info struct SignerInfo = new Win32.CMSG_SIGNER_ENCODE_INFO(); SignerInfo.cbSize = Marshal.SizeOf(SignerInfo); SignerInfo.pCertInfo = CertContexts[0].pCertInfo; SignerInfo.hCryptProvOrhNCryptKey = hProv; SignerInfo.dwKeySpec = (int)key.CspKeyContainerInfo.KeyNumber; SignerInfo.HashAlgorithm.pszObjId = Win32.szOID_OIWSEC_sha1; // Populate Signed Info struct SignedInfo = new Win32.CMSG_SIGNED_ENCODE_INFO(); SignedInfo.cbSize = Marshal.SizeOf(SignedInfo); SignedInfo.cSigners = 1; SignerInfoPtr = Marshal.AllocHGlobal(Marshal.SizeOf(SignerInfo)); Marshal.StructureToPtr(SignerInfo, SignerInfoPtr, false); SignedInfo.rgSigners = SignerInfoPtr; SignedInfo.cCertEncoded = CertBlobs.Length; CertBlobsPtr = Marshal.AllocHGlobal(Marshal.SizeOf(CertBlobs[0]) * CertBlobs.Length); for (int i = 0; i < CertBlobs.Length; i++) { Marshal.StructureToPtr(CertBlobs[i], new IntPtr(CertBlobsPtr.ToInt64() + (Marshal.SizeOf(CertBlobs[i]) * i)), false); } SignedInfo.rgCertEncoded = CertBlobsPtr; // Populate Stream Info struct StreamInfo = new Win32.CMSG_STREAM_INFO(); StreamInfo.cbContent = dwFileSize; StreamInfo.pfnStreamOutput = new Win32.StreamOutputCallbackDelegate(StreamOutputCallback); // TODO: CMSG_DETACHED_FLAG // Open message to encode hMsg = Win32.CryptMsgOpenToEncode( Win32.X509_ASN_ENCODING | Win32.PKCS_7_ASN_ENCODING, 0, Win32.CMSG_SIGNED, ref SignedInfo, null, ref StreamInfo ); if (hMsg.Equals(IntPtr.Zero)) { throw new Exception("CryptMsgOpenToEncode error #" + Marshal.GetLastWin32Error().ToString(), new Win32Exception(Marshal.GetLastWin32Error())); } // Process the whole message gchandle = GCHandle.Alloc(pbData, GCHandleType.Pinned); pbPtr = gchandle.AddrOfPinnedObject(); dwRemaining = dwFileSize; dwSize = (dwFileSize < 1024 * 1000 * 100) ? dwFileSize : 1024 * 1000 * 100; while (dwRemaining > 0) { // Update message piece by piece bResult = Win32.CryptMsgUpdate( hMsg, pbPtr, dwSize, (dwRemaining <= dwSize) ? true : false ); if (!bResult) { throw new Exception("CryptMsgUpdate error #" + Marshal.GetLastWin32Error().ToString(), new Win32Exception(Marshal.GetLastWin32Error())); } // Move to the next piece pbPtr = new IntPtr(pbPtr.ToInt64() + dwSize); dwRemaining -= dwSize; if (dwRemaining < dwSize) { dwSize = dwRemaining; } } } finally { // Clean up if (gchandle.IsAllocated) { gchandle.Free(); } if (stream != null) { stream.Close(); } if (m_callbackFile != null) { m_callbackFile.Close(); } if (!CertBlobsPtr.Equals(IntPtr.Zero)) { Marshal.FreeHGlobal(CertBlobsPtr); } if (!SignerInfoPtr.Equals(IntPtr.Zero)) { Marshal.FreeHGlobal(SignerInfoPtr); } if (!hProv.Equals(IntPtr.Zero)) { Win32.CryptReleaseContext(hProv, 0); } if (!hMsg.Equals(IntPtr.Zero)) { Win32.CryptMsgClose(hMsg); } } } // Decode CMS with streaming to support large data public void Decode(FileStream inFile, FileStream outFile) { // Variables Win32.CMSG_STREAM_INFO StreamInfo; Win32.CERT_CONTEXT SignerCertContext; BinaryReader stream = null; GCHandle gchandle = new GCHandle(); IntPtr hMsg = IntPtr.Zero; IntPtr pSignerCertInfo = IntPtr.Zero; IntPtr pSignerCertContext = IntPtr.Zero; IntPtr pbPtr = IntPtr.Zero; IntPtr hStore = IntPtr.Zero; Byte[] pbData; Boolean bResult = false; int dwFileSize; int dwRemaining; int dwSize; int cbSignerCertInfo; try { // Get data to decode dwFileSize = (int)inFile.Length; stream = new BinaryReader(inFile); pbData = stream.ReadBytes(dwFileSize); // Prepare stream for decoded info m_callbackFile = outFile; // Populate Stream Info struct StreamInfo = new Win32.CMSG_STREAM_INFO(); StreamInfo.cbContent = dwFileSize; StreamInfo.pfnStreamOutput = new Win32.StreamOutputCallbackDelegate(StreamOutputCallback); // Open message to decode hMsg = Win32.CryptMsgOpenToDecode( Win32.X509_ASN_ENCODING | Win32.PKCS_7_ASN_ENCODING, 0, 0, IntPtr.Zero, IntPtr.Zero, ref StreamInfo ); if (hMsg.Equals(IntPtr.Zero)) { throw new Exception("CryptMsgOpenToDecode error #" + Marshal.GetLastWin32Error().ToString(), new Win32Exception(Marshal.GetLastWin32Error())); } // Process the whole message gchandle = GCHandle.Alloc(pbData, GCHandleType.Pinned); pbPtr = gchandle.AddrOfPinnedObject(); dwRemaining = dwFileSize; dwSize = (dwFileSize < 1024 * 1000 * 100) ? dwFileSize : 1024 * 1000 * 100; while (dwRemaining > 0) { // Update message piece by piece bResult = Win32.CryptMsgUpdate( hMsg, pbPtr, dwSize, (dwRemaining <= dwSize) ? true : false ); if (!bResult) { throw new Exception("CryptMsgUpdate error #" + Marshal.GetLastWin32Error().ToString(), new Win32Exception(Marshal.GetLastWin32Error())); } // Move to the next piece pbPtr = new IntPtr(pbPtr.ToInt64() + dwSize); dwRemaining -= dwSize; if (dwRemaining < dwSize) { dwSize = dwRemaining; } } // Get signer certificate info cbSignerCertInfo = 0; bResult = Win32.CryptMsgGetParam( hMsg, Win32.CMSG_SIGNER_CERT_INFO_PARAM, 0, IntPtr.Zero, ref cbSignerCertInfo ); if (!bResult) { throw new Exception("CryptMsgGetParam error #" + Marshal.GetLastWin32Error().ToString(), new Win32Exception(Marshal.GetLastWin32Error())); } pSignerCertInfo = Marshal.AllocHGlobal(cbSignerCertInfo); bResult = Win32.CryptMsgGetParam( hMsg, Win32.CMSG_SIGNER_CERT_INFO_PARAM, 0, pSignerCertInfo, ref cbSignerCertInfo ); if (!bResult) { throw new Exception("CryptMsgGetParam error #" + Marshal.GetLastWin32Error().ToString(), new Win32Exception(Marshal.GetLastWin32Error())); } // Open a cert store in memory with the certs from the message hStore = Win32.CertOpenStore( Win32.CERT_STORE_PROV_MSG, Win32.X509_ASN_ENCODING | Win32.PKCS_7_ASN_ENCODING, IntPtr.Zero, 0, hMsg ); if (hStore.Equals(IntPtr.Zero)) { throw new Exception("CertOpenStore error #" + Marshal.GetLastWin32Error().ToString(), new Win32Exception(Marshal.GetLastWin32Error())); } // Find the signer's cert in the store pSignerCertContext = Win32.CertGetSubjectCertificateFromStore( hStore, Win32.X509_ASN_ENCODING | Win32.PKCS_7_ASN_ENCODING, pSignerCertInfo ); if (pSignerCertContext.Equals(IntPtr.Zero)) { throw new Exception("CertGetSubjectCertificateFromStore error #" + Marshal.GetLastWin32Error().ToString(), new Win32Exception(Marshal.GetLastWin32Error())); } // Set message for verifying SignerCertContext = (Win32.CERT_CONTEXT)Marshal.PtrToStructure(pSignerCertContext, typeof(Win32.CERT_CONTEXT)); bResult = Win32.CryptMsgControl( hMsg, 0, Win32.CMSG_CTRL_VERIFY_SIGNATURE, SignerCertContext.pCertInfo ); if (!bResult) { throw new Exception("CryptMsgControl error #" + Marshal.GetLastWin32Error().ToString(), new Win32Exception(Marshal.GetLastWin32Error())); } } finally { // Clean up if (gchandle.IsAllocated) { gchandle.Free(); } if (!pSignerCertContext.Equals(IntPtr.Zero)) { Win32.CertFreeCertificateContext(pSignerCertContext); } if (!pSignerCertInfo.Equals(IntPtr.Zero)) { Marshal.FreeHGlobal(pSignerCertInfo); } if (!hStore.Equals(IntPtr.Zero)) { Win32.CertCloseStore(hStore, Win32.CERT_CLOSE_STORE_FORCE_FLAG); } if (stream != null) { stream.Close(); } if (m_callbackFile != null) { m_callbackFile.Close(); } if (!hMsg.Equals(IntPtr.Zero)) { Win32.CryptMsgClose(hMsg); } } } } } and using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Runtime.InteropServices; using System.Security.Cryptography.X509Certificates; using System.ComponentModel; using System.Security.Cryptography; namespace LargeCMS { class Win32 { #region "CONSTS" public const int X509_ASN_ENCODING = 0x00000001; public const int PKCS_7_ASN_ENCODING = 0x00010000; public const int CMSG_SIGNED = 2; public const int CMSG_DETACHED_FLAG = 0x00000004; public const int AT_KEYEXCHANGE = 1; public const int AT_SIGNATURE = 2; public const String szOID_OIWSEC_sha1 = "1.3.14.3.2.26"; public const int CMSG_CTRL_VERIFY_SIGNATURE = 1; public const int CMSG_CERT_PARAM = 12; public const int CMSG_SIGNER_CERT_INFO_PARAM = 7; public const int CERT_STORE_PROV_MSG = 1; public const int CERT_CLOSE_STORE_FORCE_FLAG = 1; #endregion #region "STRUCTS" [StructLayout(LayoutKind.Sequential)] public struct CRYPT_ALGORITHM_IDENTIFIER { public String pszObjId; BLOB Parameters; } [StructLayout(LayoutKind.Sequential)] public struct CERT_ID { public int dwIdChoice; public BLOB IssuerSerialNumberOrKeyIdOrHashId; } [StructLayout(LayoutKind.Sequential)] public struct CMSG_SIGNER_ENCODE_INFO { public int cbSize; public IntPtr pCertInfo; public IntPtr hCryptProvOrhNCryptKey; public int dwKeySpec; public CRYPT_ALGORITHM_IDENTIFIER HashAlgorithm; public IntPtr pvHashAuxInfo; public int cAuthAttr; public IntPtr rgAuthAttr; public int cUnauthAttr; public IntPtr rgUnauthAttr; public CERT_ID SignerId; public CRYPT_ALGORITHM_IDENTIFIER HashEncryptionAlgorithm; public IntPtr pvHashEncryptionAuxInfo; } [StructLayout(LayoutKind.Sequential)] public struct CERT_CONTEXT { public int dwCertEncodingType; public IntPtr pbCertEncoded; public int cbCertEncoded; public IntPtr pCertInfo; public IntPtr hCertStore; } [StructLayout(LayoutKind.Sequential)] public struct BLOB { public int cbData; public IntPtr pbData; } [StructLayout(LayoutKind.Sequential)] public struct CMSG_SIGNED_ENCODE_INFO { public int cbSize; public int cSigners; public IntPtr rgSigners; public int cCertEncoded; public IntPtr rgCertEncoded; public int cCrlEncoded; public IntPtr rgCrlEncoded; public int cAttrCertEncoded; public IntPtr rgAttrCertEncoded; } [StructLayout(LayoutKind.Sequential)] public struct CMSG_STREAM_INFO { public int cbContent; public StreamOutputCallbackDelegate pfnStreamOutput; public IntPtr pvArg; } #endregion #region "DELEGATES" public delegate Boolean StreamOutputCallbackDelegate(IntPtr pvArg, IntPtr pbData, int cbData, Boolean fFinal); #endregion #region "API" [DllImport("advapi32.dll", CharSet = CharSet.Auto, SetLastError = true)] public static extern Boolean CryptAcquireContext( ref IntPtr hProv, String pszContainer, String pszProvider, int dwProvType, int dwFlags ); [DllImport("Crypt32.dll", SetLastError = true)] public static extern IntPtr CryptMsgOpenToEncode( int dwMsgEncodingType, int dwFlags, int dwMsgType, ref CMSG_SIGNED_ENCODE_INFO pvMsgEncodeInfo, String pszInnerContentObjID, ref CMSG_STREAM_INFO pStreamInfo ); [DllImport("Crypt32.dll", SetLastError = true)] public static extern IntPtr CryptMsgOpenToDecode( int dwMsgEncodingType, int dwFlags, int dwMsgType, IntPtr hCryptProv, IntPtr pRecipientInfo, ref CMSG_STREAM_INFO pStreamInfo ); [DllImport("Crypt32.dll", SetLastError = true)] public static extern Boolean CryptMsgClose( IntPtr hCryptMsg ); [DllImport("Crypt32.dll", SetLastError = true)] public static extern Boolean CryptMsgUpdate( IntPtr hCryptMsg, Byte[] pbData, int cbData, Boolean fFinal ); [DllImport("Crypt32.dll", SetLastError = true)] public static extern Boolean CryptMsgUpdate( IntPtr hCryptMsg, IntPtr pbData, int cbData, Boolean fFinal ); [DllImport("Crypt32.dll", SetLastError = true)] public static extern Boolean CryptMsgGetParam( IntPtr hCryptMsg, int dwParamType, int dwIndex, IntPtr pvData, ref int pcbData ); [DllImport("Crypt32.dll", SetLastError = true)] public static extern Boolean CryptMsgControl( IntPtr hCryptMsg, int dwFlags, int dwCtrlType, IntPtr pvCtrlPara ); [DllImport("advapi32.dll", SetLastError = true)] public static extern Boolean CryptReleaseContext( IntPtr hProv, int dwFlags ); [DllImport("Crypt32.dll", SetLastError = true)] public static extern IntPtr CertCreateCertificateContext( int dwCertEncodingType, IntPtr pbCertEncoded, int cbCertEncoded ); [DllImport("Crypt32.dll", SetLastError = true)] public static extern Boolean CertFreeCertificateContext( IntPtr pCertContext ); [DllImport("Crypt32.dll", SetLastError = true)] public static extern IntPtr CertOpenStore( int lpszStoreProvider, int dwMsgAndCertEncodingType, IntPtr hCryptProv, int dwFlags, IntPtr pvPara ); [DllImport("Crypt32.dll", SetLastError = true)] public static extern IntPtr CertGetSubjectCertificateFromStore( IntPtr hCertStore, int dwCertEncodingType, IntPtr pCertId ); [DllImport("Crypt32.dll", SetLastError = true)] public static extern IntPtr CertCloseStore( IntPtr hCertStore, int dwFlags ); #endregion } }

  Read the article

• How to optimize dynamic programming?

  - by Chan

  Problem A number is called lucky if the sum of its digits, as well as the sum of the squares of its digits is a prime number. How many numbers between A and B are lucky? Input: The first line contains the number of test cases T. Each of the next T lines contains two integers, A and B. Output: Output T lines, one for each case containing the required answer for the corresponding case. Constraints: 1 <= T <= 10000 1 <= A <= B <= 10^18 Sample Input: 2 1 20 120 130 Sample Output: 4 1 Explanation: For the first case, the lucky numbers are 11, 12, 14, 16. For the second case, the only lucky number is 120. The problem is quite simple if we use brute force, however the running time is so critical that my program failed most test cases. My current idea is to use dynamic programming by storing the previous sum in a temporary array, so for example: sum_digits(10) = 1 -> sum_digits(11) = sum_digits(10) + 1 The same idea is applied for sum square but with counter equals to odd numbers. Unfortunately, it still failed 9 of 10 test cases which makes me think there must be a better way to solve it. Any idea would be greatly appreciated. #include <iostream> #include <vector> #include <string> #include <algorithm> #include <unordered_map> #include <unordered_set> #include <cmath> #include <cassert> #include <bitset> using namespace std; bool prime_table[1540] = { 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0 }; unsigned num_digits(long long i) { return i > 0 ? (long) log10 ((double) i) + 1 : 1; } void get_sum_and_sum_square_digits(long long n, int& sum, int& sum_square) { sum = 0; sum_square = 0; int digit; while (n) { digit = n % 10; sum += digit; sum_square += digit * digit; n /= 10; } } void init_digits(long long n, long long previous_sum[], const int size = 18) { int current_no_digits = num_digits(n); int digit; for (int i = 0; i < current_no_digits; ++i) { digit = n % 10; previous_sum[i] = digit; n /= 10; } for (int i = current_no_digits; i <= size; ++i) { previous_sum[i] = 0; } } void display_previous(long long previous[]) { for (int i = 0; i < 18; ++i) { cout << previous[i] << ","; } } int count_lucky_number(long long A, long long B) { long long n = A; long long end = B; int sum = 0; int sum_square = 0; int lucky_counter = 0; get_sum_and_sum_square_digits(n, sum, sum_square); long long sum_counter = sum; long long sum_square_counter = sum_square; if (prime_table[sum_counter] && prime_table[sum_square_counter]) { lucky_counter++; } long long previous_sum[19] = {1}; init_digits(n, previous_sum); while (n < end) { n++; if (n % 100000000000000000 == 0) { previous_sum[17]++; sum_counter = previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[16] = 0; previous_sum[15] = 0; previous_sum[14] = 0; previous_sum[13] = 0; previous_sum[12] = 0; previous_sum[11] = 0; previous_sum[10] = 0; previous_sum[9] = 0; previous_sum[8] = 0; previous_sum[7] = 0; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 10000000000000000 == 0) { previous_sum[16]++; sum_counter = previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[15] = 0; previous_sum[14] = 0; previous_sum[13] = 0; previous_sum[12] = 0; previous_sum[11] = 0; previous_sum[10] = 0; previous_sum[9] = 0; previous_sum[8] = 0; previous_sum[7] = 0; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 1000000000000000 == 0) { previous_sum[15]++; sum_counter = previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[14] = 0; previous_sum[13] = 0; previous_sum[12] = 0; previous_sum[11] = 0; previous_sum[10] = 0; previous_sum[9] = 0; previous_sum[8] = 0; previous_sum[7] = 0; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 100000000000000 == 0) { previous_sum[14]++; sum_counter = previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[13] = 0; previous_sum[12] = 0; previous_sum[11] = 0; previous_sum[10] = 0; previous_sum[9] = 0; previous_sum[8] = 0; previous_sum[7] = 0; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 10000000000000 == 0) { previous_sum[13]++; sum_counter = previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[12] = 0; previous_sum[11] = 0; previous_sum[10] = 0; previous_sum[9] = 0; previous_sum[8] = 0; previous_sum[7] = 0; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 1000000000000 == 0) { previous_sum[12]++; sum_counter = previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[11] = 0; previous_sum[10] = 0; previous_sum[9] = 0; previous_sum[8] = 0; previous_sum[7] = 0; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 100000000000 == 0) { previous_sum[11]++; sum_counter = previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[11] * previous_sum[11] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[10] = 0; previous_sum[9] = 0; previous_sum[8] = 0; previous_sum[7] = 0; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 10000000000 == 0) { previous_sum[10]++; sum_counter = previous_sum[10] + previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[10] * previous_sum[10] + previous_sum[11] * previous_sum[11] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[9] = 0; previous_sum[8] = 0; previous_sum[7] = 0; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 1000000000 == 0) { previous_sum[9]++; sum_counter = previous_sum[9] + previous_sum[10] + previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[9] * previous_sum[9] + previous_sum[10] * previous_sum[10] + previous_sum[11] * previous_sum[11] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[8] = 0; previous_sum[7] = 0; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 100000000 == 0) { previous_sum[8]++; sum_counter = previous_sum[8] + previous_sum[9] + previous_sum[10] + previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[8] * previous_sum[8] + previous_sum[9] * previous_sum[9] + previous_sum[10] * previous_sum[10] + previous_sum[11] * previous_sum[11] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[7] = 0; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 10000000 == 0) { previous_sum[7]++; sum_counter = previous_sum[7] + previous_sum[8] + previous_sum[9] + previous_sum[10] + previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[7] * previous_sum[7] + previous_sum[8] * previous_sum[8] + previous_sum[9] * previous_sum[9] + previous_sum[10] * previous_sum[10] + previous_sum[11] * previous_sum[11] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[6] = 0; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 1000000 == 0) { previous_sum[6]++; sum_counter = previous_sum[6] + previous_sum[7] + previous_sum[8] + previous_sum[9] + previous_sum[10] + previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[6] * previous_sum[6] + previous_sum[7] * previous_sum[7] + previous_sum[8] * previous_sum[8] + previous_sum[9] * previous_sum[9] + previous_sum[10] * previous_sum[10] + previous_sum[11] * previous_sum[11] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[5] = 0; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 100000 == 0) { previous_sum[5]++; sum_counter = previous_sum[5] + previous_sum[6] + previous_sum[7] + previous_sum[8] + previous_sum[9] + previous_sum[10] + previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[5] * previous_sum[5] + previous_sum[6] * previous_sum[6] + previous_sum[7] * previous_sum[7] + previous_sum[8] * previous_sum[8] + previous_sum[9] * previous_sum[9] + previous_sum[10] * previous_sum[10] + previous_sum[11] * previous_sum[11] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[4] = 0; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 10000 == 0) { previous_sum[4]++; sum_counter = previous_sum[4] + previous_sum[5] + previous_sum[6] + previous_sum[7] + previous_sum[8] + previous_sum[9] + previous_sum[10] + previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[4] * previous_sum[4] + previous_sum[5] * previous_sum[5] + previous_sum[6] * previous_sum[6] + previous_sum[7] * previous_sum[7] + previous_sum[8] * previous_sum[8] + previous_sum[9] * previous_sum[9] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[3] = 0; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 1000 == 0) { previous_sum[3]++; sum_counter = previous_sum[3] + previous_sum[4] + previous_sum[5] + previous_sum[6] + previous_sum[7] + previous_sum[8] + previous_sum[9] + previous_sum[10] + previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[3] * previous_sum[3] + previous_sum[4] * previous_sum[4] + previous_sum[5] * previous_sum[5] + previous_sum[6] * previous_sum[6] + previous_sum[7] * previous_sum[7] + previous_sum[8] * previous_sum[8] + previous_sum[9] * previous_sum[9] + previous_sum[10] * previous_sum[10] + previous_sum[11] * previous_sum[11] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[2] = 0; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 100 == 0) { previous_sum[2]++; sum_counter = previous_sum[2] + previous_sum[3] + previous_sum[4] + previous_sum[5] + previous_sum[6] + previous_sum[7] + previous_sum[8] + previous_sum[9] + previous_sum[10] + previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[2] * previous_sum[2] + previous_sum[3] * previous_sum[3] + previous_sum[4] * previous_sum[4] + previous_sum[5] * previous_sum[5] + previous_sum[6] * previous_sum[6] + previous_sum[7] * previous_sum[7] + previous_sum[8] * previous_sum[8] + previous_sum[9] * previous_sum[9] + previous_sum[10] * previous_sum[10] + previous_sum[11] * previous_sum[11] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[1] = 0; previous_sum[0] = 0; } else if (n % 10 == 0) { previous_sum[1]++; sum_counter = previous_sum[1] + previous_sum[2] + previous_sum[3] + previous_sum[4] + previous_sum[5] + previous_sum[6] + previous_sum[7] + previous_sum[8] + previous_sum[9] + previous_sum[10] + previous_sum[11] + previous_sum[12] + previous_sum[13] + previous_sum[14] + previous_sum[15] + previous_sum[16] + previous_sum[17] + previous_sum[18]; sum_square_counter = previous_sum[1] * previous_sum[1] + previous_sum[2] * previous_sum[2] + previous_sum[3] * previous_sum[3] + previous_sum[4] * previous_sum[4] + previous_sum[5] * previous_sum[5] + previous_sum[6] * previous_sum[6] + previous_sum[7] * previous_sum[7] + previous_sum[8] * previous_sum[8] + previous_sum[9] * previous_sum[9] + previous_sum[10] * previous_sum[10] + previous_sum[11] * previous_sum[11] + previous_sum[12] * previous_sum[12] + previous_sum[13] * previous_sum[13] + previous_sum[14] * previous_sum[14] + previous_sum[15] * previous_sum[15] + previous_sum[16] * previous_sum[16] + previous_sum[17] * previous_sum[17] + previous_sum[18] * previous_sum[18]; previous_sum[0] = 0; } else { sum_counter++; sum_square_counter += ((n - 1) % 10) * 2 + 1; } // get_sum_and_sum_square_digits(n, sum, sum_square); // assert(sum == sum_counter && sum_square == sum_square_counter); if (prime_table[sum_counter] && prime_table[sum_square_counter]) { lucky_counter++; } } return lucky_counter; } void inout_lucky_numbers() { int n; cin >> n; long long a; long long b; while (n--) { cin >> a >> b; cout << count_lucky_number(a, b) << endl; } } int main() { inout_lucky_numbers(); return 0; }

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• Find kth smallest element in a binary search tree in Optimum way

  - by Bragaadeesh

  Hi, I need to find the kth smallest element in the binary search tree without using any static/global variable. How to achieve it efficiently? The solution that I have in my mind is doing the operation in O(n), the worst case since I am planning to do an inorder traversal of the entire tree. But deep down I feel that I am not using the BST property here. Is my assumptive solution correct or is there a better one available ?

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• Write a function that compares two strings and returns a third string containing only the letters th

  - by Pritam

  Hi All, I got this homework. And have solved it in following way. I need your comments whether it is a good approach or I need to use any other data sturcture to solve it in better way. public string ReturnCommon(string firstString, string scndString) { StringBuilder newStb = new StringBuilder(); if (firstString != null && scndString != null) { foreach (char ichar in firstString) { if (!newStb.ToString().Contains(ichar) && scndString.Contains(ichar)) newStb.Append(ichar); } } return newStb.ToString(); }

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• Point in polygon OR point on polygon using LINQ

  - by wageoghe

  As noted in an earlier question, How to Zip enumerable with itself, I am working on some math algorithms based on lists of points. I am currently working on point in polygon. I have the code for how to do that and have found several good references here on SO, such as this link Hit test. So, I can figure out whether or not a point is in a polygon. As part of determining that, I want to determine if the point is actually on the polygon. This I can also do. If I can do all of that, what is my question you might ask? Can I do it efficiently using LINQ? I can already do something like the following (assuming a Pairwise extension method as described in my earlier question as well as in links to which my question/answers links, and assuming a Position type that has X and Y members). I have not tested much, so the lambda might not be 100% correct. Also, it does not take very small differences into account. public static PointInPolygonLocation PointInPolygon(IEnumerable<Position> pts, Position pt) { int numIntersections = pts.Pairwise( (p1, p2) => { if (p1.Y != p2.Y) { if ((p1.Y >= pt.Y && p2.Y < pt.Y) || (p1.Y < pt.Y && p2.Y >= pt.Y)) { if (p1.X < p1.X && p2.X < pt.X) { return 1; } if (p1.X < pt.X || p2.X < pt.X) { if (((pt.Y - p1.Y) * ((p1.X - p2.X) / (p1.Y - p2.Y)) * p1.X) < pt.X) { return 1; } } } } return 0; }).Sum(); if (numIntersections % 2 == 0) { return PointInPolygonLocation.Outside; } else { return PointInPolygonLocation.Inside; } } This function, PointInPolygon, takes the input Position, pt, iterates over the input sequence of position values, and uses the Jordan Curve method to determine how many times a ray extended from pt to the left intersects the polygon. The lambda expression will yield, into the "zipped" list, 1 for every segment that is crossed, and 0 for the rest. The sum of these values determines if pt is inside or outside of the polygon (odd == inside, even == outside). So far, so good. Now, for any consecutive pairs of position values in the sequence (i.e. in any execution of the lambda), we can also determine if pt is ON the segment p1, p2. If that is the case, we can stop the calculation because we have our answer. Ultimately, my question is this: Can I perform this calculation (maybe using Aggregate?) such that we will only iterate over the sequence no more than 1 time AND can we stop the iteration if we encounter a segment that pt is ON? In other words, if pt is ON the very first segment, there is no need to examine the rest of the segments because we have the answer. It might very well be that this operation (particularly the requirement/desire to possibly stop the iteration early) does not really lend itself well to the LINQ approach. It just occurred to me that maybe the lambda expression could yield a tuple, the intersection value (1 or 0 or maybe true or false) and the "on" value (true or false). Maybe then I could use TakeWhile(anontype.PointOnPolygon == false). If I Sum the tuples and if ON == 1, then the point is ON the polygon. Otherwise, the oddness or evenness of the sum of the other part of the tuple tells if the point is inside or outside.

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• Best practice to calculate the average speed from GPS coordinates

  - by Sebi

  i have here a device which can give me gps coordinates. the time intervall i can define. i want to use it to calculate the average speed during driving or travelling by car. actually i used a orthodrome formula to calculate the distance between two points and then divided it by the given time intervall. by the implemenation i followed this term (http://de.wikipedia.org/wiki/Orthodrome#Genauere_Formel_zur_Abstandsberechnung_auf_der_Erde). Unfortunately i could only find a german link, but i think the formula should be understandable in any language ;) Unfortunately, using this formula and a time intverall of 1 seconds gives very unprecises results. the speed while walking is between 1 km/h and 20km/h. So i wonder if there is a general reference how to implement distance calculation between two gps coordinates (i found something similar on SO) and particulary, which is the best time intervall to update the GPS coordiantes.

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• How do I detect overflow while multiplying two 2's complement integers?

  - by eSKay

  I want to multiply two numbers, and detect if there was an overflow. What is the simplest way to do that?

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• Dynamic programming - Coin change decision problem?

  - by Tony

  I'm reviewing some old notes from my algorithms course and the dynamic programming problems are seeming a bit tricky to me. I have a problem where we have an unlimited supply of coins, with some denominations x1, x2, ... xn and we want to make change for some value X. We are trying to design a dynamic program to decide whether change for X can be made or not (not minimizing the number of coins, or returning which coins, just true or false). I've done some thinking about this problem, and I can see a recursive method of doing this where it's something like... MakeChange(X, x[1..n this is the coins]) for (int i = 1; i < n; i++) { if ( (X - x[i] ==0) || MakeChange(X - x[i]) ) return true; } return false; Converting this a dynamic program is not coming so easily to me. How might I approach this?

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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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• Find the "largest" dense sub matrix in a large sparse matrix

  - by BCS

  Given a large sparse matrix (say 10k+ by 1M+) I need to find a subset, not necessarily continuous, of the rows and columns that form a dense matrix (all non-zero elements). I want this sub matrix to be as large as possible (not the largest sum, but the largest number of elements) within some aspect ratio constraints. Are there any known exact or aproxamate solutions to this problem? A quick scan on Google seems to give a lot of close-but-not-exactly results. What terms should I be looking for? edit: Just to clarify; the sub matrix need not be continuous. In fact the row and column order is completely arbitrary so adjacency is completely irrelevant. A thought based on Chad Okere's idea Order the rows from largest count to smallest count (not necessary but might help perf) Select two rows that have a "large" overlap Add all other rows that won't reduce the overlap Record that set Add whatever row reduces the overlap by the least Repeat at #3 until the result gets to small Start over at #2 with a different starting pair Continue until you decide the result is good enough

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