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  • how to bind the same vector multiple times using R?

    - by hendrik
    question: how can i bind the same vector, lets say o=c(1,2,3,4) mutiple times to get a matrix like o=array(c(1,2,3,4,1,2,3,4,1,2,3,4), dim(c(4,3)) o [,1] [,2] [,3] [1,] 1 1 1 [2,] 2 2 2 [3,] 3 3 3 [4,] 4 4 4 in a nicer way then: o=cbind(o,o,o) and maybe more generalized (dublicate()?? I need this to specifiy colors for elements in textplot() thx a lot

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  • Faster method for Matrix vector product for large matrix in C or C++ for use in GMRES

    - by user35959
    I have a large, dense matrix A, and I aim to find the solution to the linear system Ax=b using an iterative method (in MATLAB was the plan using its built in GMRES). For more than 10,000 rows, this is too much for my computer to store in memory, but I know that the entries in A are constructed by two known vectors x and y of length N and the entries satisfy: A(i,j) = .5*(x[i]-x[j])^2+([y[i]-y[j])^2 * log(x[i]-x[j])^2+([y[i]-y[j]^2). MATLAB's GMRES command accepts as input a function call that can compute the matrix vector product A*x, which allows me to handle larger matrices than I can store in memory. To write the matrix-vecotr product function, I first tried this in matlab by going row by row and using some vectorization, but I avoid spawning the entire array A (since it would be too large). This was fairly slow unfortnately in my application for GMRES. My plan was to write a mex file for MATLAB to, which is in C, and ideally should be significantly faster than the matlab code. I'm rather new to C, so this went rather poorly and my naive attempt at writing the code in C was slower than my partially vectorized attempt in Matlab. #include <math.h> #include "mex.h" void Aproduct(double *x, double *ctrs_x, double *ctrs_y, double *b, mwSize n) { mwSize i; mwSize j; double val; for (i=0; i<n; i++) { for (j=0; j<i; j++) { val = pow(ctrs_x[i]-ctrs_x[j],2)+pow(ctrs_y[i]-ctrs_y[j],2); b[i] = b[i] + .5* val * log(val) * x[j]; } for (j=i+1; j<n; j++) { val = pow(ctrs_x[i]-ctrs_x[j],2)+pow(ctrs_y[i]-ctrs_y[j],2); b[i] = b[i] + .5* val * log(val) * x[j]; } } } The above is the computational portion of the code for the matlab mex file (which is slightly modified C, if I understand correctly). Please note that I skip the case i=j, since in that case the variable val will be a 0*log(0), which should be interpreted as 0 for me, so I just skip it. Is there a more efficient or faster way to write this? When I call this C function via the mex file in matlab, it is quite slow, slower even than the matlab method I used. This surprises me since I suspected that C code should be much faster than matlab. The alternative matlab method which is partially vectorized that I am comparing it with is function Ax = Aprod(x,ctrs) n = length(x); Ax = zeros(n,1); for j=1:(n-3) v = .5*((ctrs(j,1)-ctrs(:,1)).^2+(ctrs(j,2)-ctrs(:,2)).^2).*log((ctrs(j,1)-ctrs(:,1)).^2+(ctrs(j,2)-ctrs(:,2)).^2); v(j)=0; Ax(j) = dot(v,x(1:n-3); end (the n-3 is because there is actually 3 extra components, but they are dealt with separately,so I excluded that code). This is partly vectorized and only needs one for loop, so it makes some sense that it is faster. However, I was hoping I could go even faster with C+mex file. Any suggestions or help would be greatly appreciated! Thanks! EDIT: I should be more clear. I am open to any faster method that can help me use GMRES to invert this matrix that I am interested in, which requires a faster way of doing the matrix vector product without explicitly loading the array into memory. Thanks!

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  • Exercise 26 of The Pragmatic Programmer

    - by _ande_turner_
    There is a code snippet presented in The Pragmatic Programmer on page 143 as: public class Colada { private Blender myBlender; private Vector myStuff; public Colada() { myBlender = new Blender(); myStuff = new Vector(); } private doSomething() { myBlender.addIngredients(myStuff.elements()); } } This obeys the Law of Demeter / Principle of Least Knowledge. Is it preferable to, and are there any caveats for, replacing it with the following, which utilises Dependency Injection? public class Colada throws IllegalArgumentException { private Blender myBlender; private Vector myStuff; public Colada(Blender blender, Vector stuff) { blender == null ? throw new IllegalArgumentException() : myBlender = blender; stuff == null ? throw new IllegalArgumentException() : myStuff = stuff; } public getInstance() { Blender blender = new Blender(); Vector stuff = new Vector(); return new Colada(blender, stuff); } private doSomething() { myBlender.addIngredients(myStuff.elements()); } }

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  • Change the size of the text in legend according to the length of the legend vector in the graph

    - by user1021713
    I have to draw a 20 plots and horizontally place a legends in each plots. I gave the following command for the first plot: plot(x=1:4,y=1:4) legend("bottom",legend = c("a","b","c","d"),horiz=TRUE,text.font=2,cex=0.64) then for the second plot I tried : plot(x=1:2,y=1:2) legend("bottom",legend = c("a","b"),horiz=TRUE,text.font=2,cex=0.64) But because the size of the character vector passed to legend argument are different I get the size of the legend different. Since I have to plot so many different plots having varying sizes of legends,I would want to do it in an automated fashion. Is there a way to do this which can fix the size of the legend in all the plots and fit it to graph size?

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  • Problem in running boost eample blocking_udp_echo_client on MacOSX

    - by n179911
    I am trying to run blocking_udp_echo_client on MacOS X http://www.boost.org/doc/libs/1_35_0/doc/html/boost_asio/example/echo/blocking_udp_echo_client.cpp I run it with argument 'localhost 9000' But the program crashes and this is the line in the source which crashes: `udp::socket s(io_service, udp::endpoint(udp::v4(), 0));' this is the stack trace: #0 0x918c3e42 in __kill #1 0x918c3e34 in kill$UNIX2003 #2 0x9193623a in raise #3 0x91942679 in abort #4 0x940d96f9 in __gnu_debug::_Error_formatter::_M_error #5 0x0000e76e in __gnu_debug::_Safe_iterator::op_base* , __gnu_debug_def::list::op_base*, std::allocator::op_base* ::_Safe_iterator at safe_iterator.h:124 #6 0x00014729 in boost::asio::detail::hash_map::op_base*::bucket_type::bucket_type at hash_map.hpp:277 #7 0x00019e97 in std::_Construct::op_base*::bucket_type, boost::asio::detail::hash_map::op_base*::bucket_type at stl_construct.h:81 #8 0x0001a457 in std::__uninitialized_fill_n_aux::op_base*::bucket_type*, __gnu_norm::vector::op_base*::bucket_type, std::allocator::op_base*::bucket_type , unsigned long, boost::asio::detail::hash_map::op_base*::bucket_type at stl_uninitialized.h:194 #9 0x0001a4e1 in std::uninitialized_fill_n::op_base*::bucket_type*, __gnu_norm::vector::op_base*::bucket_type, std::allocator::op_base*::bucket_type , unsigned long, boost::asio::detail::hash_map::op_base*::bucket_type at stl_uninitialized.h:218 #10 0x0001a509 in std::__uninitialized_fill_n_a::op_base*::bucket_type*, __gnu_norm::vector::op_base*::bucket_type, std::allocator::op_base*::bucket_type , unsigned long, boost::asio::detail::hash_map::op_base*::bucket_type, boost::asio::detail::hash_map::op_base*::bucket_type at stl_uninitialized.h:310 #11 0x0001aa34 in __gnu_norm::vector::op_base*::bucket_type, std::allocator::op_base*::bucket_type ::_M_fill_insert at vector.tcc:365 #12 0x0001acda in __gnu_norm::vector::op_base*::bucket_type, std::allocator::op_base*::bucket_type ::insert at stl_vector.h:658 #13 0x0001ad81 in __gnu_norm::vector::op_base*::bucket_type, std::allocator::op_base*::bucket_type ::resize at stl_vector.h:427 #14 0x0001ae3a in __gnu_debug_def::vector::op_base*::bucket_type, std::allocator::op_base*::bucket_type ::resize at vector:169 #15 0x0001b7be in boost::asio::detail::hash_map::op_base*::rehash at hash_map.hpp:221 #16 0x0001bbeb in boost::asio::detail::hash_map::op_base*::hash_map at hash_map.hpp:67 #17 0x0001bc74 in boost::asio::detail::reactor_op_queue::reactor_op_queue at reactor_op_queue.hpp:42 #18 0x0001bd24 in boost::asio::detail::kqueue_reactor::kqueue_reactor at kqueue_reactor.hpp:86 #19 0x0001c000 in boost::asio::detail::service_registry::use_service at service_registry.hpp:109 #20 0x0001c14d in boost::asio::use_service at io_service.ipp:195 #21 0x0001c26d in boost::asio::detail::reactive_socket_service ::reactive_socket_service at reactive_socket_service.hpp:111 #22 0x0001c344 in boost::asio::detail::service_registry::use_service at service_registry.hpp:109 #23 0x0001c491 in boost::asio::use_service at io_service.ipp:195 #24 0x0001c4d5 in boost::asio::datagram_socket_service::datagram_socket_service at datagram_socket_service.hpp:95 #25 0x0001c59e in boost::asio::detail::service_registry::use_service at service_registry.hpp:109 #26 0x0001c6eb in boost::asio::use_service at io_service.ipp:195 #27 0x0001c711 in boost::asio::basic_io_object ::basic_io_object at basic_io_object.hpp:72 #28 0x0001c783 in boost::asio::basic_socket ::basic_socket at basic_socket.hpp:108 #29 0x0001c865 in boost::asio::basic_datagram_socket ::basic_datagram_socket at basic_datagram_socket.hpp:107 #30 0x000027bc in main at main.cpp:32 This is the gdb output: (gdb) continue /Developer/SDKs/MacOSX10.5.sdk/usr/include/c++/4.0.0/debug/safe_iterator.h:127: error: attempt to copy-construct an iterator from a singular iterator. Objects involved in the operation: iterator "this" @ 0x0x100420 { type = N11__gnu_debug14_Safe_iteratorIN10__gnu_norm14_List_iteratorISt4pairIiPN5boost4asio6detail16reactor_op_queueIiE7op_baseEEEEN15__gnu_debug_def4listISB_SaISB_EEEEE (mutable iterator); state = singular; } iterator "other" @ 0x0xbfffe8a4 { type = N11__gnu_debug14_Safe_iteratorIN10__gnu_norm14_List_iteratorISt4pairIiPN5boost4asio6detail16reactor_op_queueIiE7op_baseEEEEN15__gnu_debug_def4listISB_SaISB_EEEEE (mutable iterator); state = singular; } Program received signal: “SIGABRT”. (gdb) continue Program received signal: “?”. Does someone has any idea why this example does not work on mac osx? Thank you.

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  • is it wasteful/bad design to use a vector/list where in most instances it will only have one element

    - by lucid
    is it wasteful/bad design to use a vector/list where in most instances it will only have one element? example: class dragon { ArrayList<head> = new ArrayList<head> Heads; tail Tail = new tail(); body Body = new body(); dragon() { theHead=new head(); Heads.add(theHead); } void nod() { for (int i=0;i<Heads.size();i++) { heads.get(i).GoUpAndDown(); } } } class firedragon extends dragon { } class icedragon extends dragon { } class lightningdragon extends dragon { } // 10 other one-headed dragon declarations here class hydra extends dragon { hydra() { anotherHead=new head(); for (int i=0;i<2;i++) { Heads.add(anotherHead); } } } class superhydra extends dragon { superhydra() { anotherHead=new head(); for (int i=0;i<4;i++) { Heads.add(anotherHead); } } }

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  • Where can I get PRINCE2 logo vector (eps) logo?

    - by WooYek
    Our organization did invest in PRINCE2 training and certification for our Project Managers and we would like to advertise the fact that we are certified and we use PRINCE2 project management method. Where can I get PRINCE2 logo in eps vector graphics format, usable for printed materials business cards and marketing brochures. I was looking for such logo on http://pm4success.com but with no result. PS. I know it's barely development connected, but it is a fact that PM metods and developer certification is a part of every developer life... like RSI ;) Please do not close this question.

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  • Classes, constructor and pointer class members

    - by pocoa
    I'm a bit confused about the object references. Please check the examples below: class ListHandler { public: ListHandler(vector<int> &list); private: vector<int> list; } ListHandler::ListHandler(vector<int> &list) { this->list = list; } Here I would be wasting memory right? So the right one would be: class ListHandler { public: ListHandler(vector<int>* list); private: vector<int>* list; } ListHandler::ListHandler(vector<int>* list) { this->list = list; } ListHandler::~ListHandler() { delete list; }

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  • Reducing Integer Fractions Algorithm - Solution Explanation?

    - by Andrew Tomazos - Fathomling
    This is a followup to this problem: Reducing Integer Fractions Algorithm Following is a solution to the problem from a grandmaster: #include <cstdio> #include <algorithm> #include <functional> using namespace std; const int MAXN = 100100; const int MAXP = 10001000; int p[MAXP]; void init() { for (int i = 2; i < MAXP; ++i) { if (p[i] == 0) { for (int j = i; j < MAXP; j += i) { p[j] = i; } } } } void f(int n, vector<int>& a, vector<int>& x) { a.resize(n); vector<int>(MAXP, 0).swap(x); for (int i = 0; i < n; ++i) { scanf("%d", &a[i]); for (int j = a[i]; j > 1; j /= p[j]) { ++x[p[j]]; } } } void g(const vector<int>& v, vector<int> w) { for (int i: v) { for (int j = i; j > 1; j /= p[j]) { if (w[p[j]] > 0) { --w[p[j]]; i /= p[j]; } } printf("%d ", i); } puts(""); } int main() { int n, m; vector<int> a, b, x, y, z; init(); scanf("%d%d", &n, &m); f(n, a, x); f(m, b, y); printf("%d %d\n", n, m); transform(x.begin(), x.end(), y.begin(), insert_iterator<vector<int> >(z, z.end()), [](int a, int b) { return min(a, b); }); g(a, z); g(b, z); return 0; } It isn't clear to me how it works. Can anyone explain it? The equivilance is as follows: a is the numerator vector of length n b is the denominator vector of length m

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  • XSS as attack vector even if XSS data not stored?

    - by Klaas van Schelven
    I have a question about XSS Can forms be used as a vector for XSS even if the data is not stored in the database and used at a later point? i.e. in php the code would be this: <form input="text" value="<?= @$_POST['my_field'] ?>" name='my_field'> Showing an alert box (demonstrate that JS can be run) on your own browser is trivial with the code above. But is this exploitable across browsers as well? The only scenario I see is where you trick someone into visiting a certain page, i.e. a combination of CSRF and XSS. "Stored in a database and used at a later point": the scenario I understand about CSS is where you're able to post data to a site that runs JavaScript and is shown on a page in a browser that has greater/different privileges than your own. But, to be clear, this is not wat I'm talking about above.

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  • custom list field with more than one row(image and text), and can be populated using a vector

    - by Jisson
    In my app I want show a list,and when user click in any list item ,some new items must be added to the same list.Now I use default ListField of bb and a vetcor to add or remove elements to or from it.Now I want the list to hold more than one row(an image and some text), i searched for it I got some good sample but it not suitable for me(http://stackoverflow.com/questions/1872160/how-to-customize-list-field-in-blackberry). I need a custom listfield which have atleast two rows and can be populated using a vector from the MainScreen class.I want only one object of custom class.If any one have idea please help.

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  • C++ - Efficient way to iterate over the contents of a vector?

    - by Francisco P.
    Hello, everyone! I am implementing a text-based version of Scrabble for a college project. I have a vector containing around 400K strings (my dictionary), and, at some point in every turn, I'm going to have to check if any word in the dictionary can be formed with the pieces in the player's hand. My only solution to this is iterating through the string, one by one, and using a sub-routine I have to check if the string in question can be formed from the player's pieces. I'll implement a quickfail checking if the user has any vowels, but it'll still be woefully inefficient. Any suggestions? Thanks for your time!

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  • C++ - How to efficiently find out if any string in a vector can be assembled from a set of letters

    - by Francisco P.
    Hello, everyone! I am implementing a text-based version of Scrabble for a college project. I have a vector containing around 400K strings (my dictionary), and, at some point in every turn, I'm going to have to check if there's still a word in the dictionary which can be formed with the pieces in the player's hand. I'm checking if the player has any move left... If not, it's game over for the player in question... My only solution to this is iterating through the string, one by one, and using a sub-routine I have to check if the string in question can be formed from the player's pieces. I'll implement a quickfail checking if the user has any vowels, but it'll still be woefully inefficient. Any suggestions? Thanks for your time!

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  • Sort vector<int>(n) in O(n) time using O(m) space?

    - by Adam
    I have a vector<unsigned int> vec of size n. Each element in vec is in the range [0, m], no duplicates, and I want to sort vec. Is it possible to do better than O(n log n) time if you're allowed to use O(m) space? In the average case m is much larger than n, in the worst case m == n. Ideally I want something O(n). I get the feeling that there's a bucket sort-ish way to do this: unsigned int aux[m]; aux[vec[i]] = i; Somehow extract the permutation and permute vec. I'm stuck on how to do 3. In my application m is on the order of 16k. However this sort is in the inner loops and accounts for a significant portion of my runtime.

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  • How to convert Vector Layer coordinates into Map Latitude and Longitude in Openlayers.

    - by Jenny
    I'm pretty confused. I have a point: x= -12669114.702301 y= 5561132.6760608 That I got from drawing a square on a vector layer with the DrawFeature controller. The numbers seem...erm...awfull large, but they seem to work, because if I later draw a square with all the same points, it's in the same position, so I figure they have to be right. The problem is when I try to convert this point to latitude and longitude. I'm using: map.getLonLatFromPixel(pointToPixel(points[0])); Where points[0] is a geometry Point, and the pointToPixel function takes any point and turns it into a pixel (since the getLonLatFromPixel needs a pixel). It does this by simply taking the point's x, and making it the pixels x, and so on. The latitude and longitude I get is on the order of: lat: -54402718463.864 lng: -18771380.353223 This is very clearly wrong. I'm left really confused. I try projecting this object, using: .transform(new OpenLayers.Projection("EPSG:4326"), map.getProjectionObject()); But I don't really get it and am pretty sure I did it incorrectly, anyways. My code is here: http://pastie.org/909644 I'm sort of at a loss. The coordinates seem consistent, because I can reuse them to get the same result...but they seem way larger than any of the examples I'm seeing on the openLayers website...

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  • How to properly cast a global memory array using the uint4 vector in CUDA to increase memory throughput?

    - by charis
    There are generally two techniques to increase the memory throughput of the global memory on a CUDA kernel; memory accesses coalescence and accessing words of at least 4 bytes. With the first technique accesses to the same memory segment by threads of the same half-warp are coalesced to fewer transactions while be accessing words of at least 4 bytes this memory segment is effectively increased from 32 bytes to 128. To access 16-byte instead of 1-byte words when there are unsigned chars stored in the global memory, the uint4 vector is commonly used by casting the memory array to uint4: uint4 *text4 = ( uint4 * ) d_text; var = text4[i]; In order to extract the 16 chars from var, i am currently using bitwise operations. For example: s_array[j * 16 + 0] = var.x & 0x000000FF; s_array[j * 16 + 1] = (var.x >> 8) & 0x000000FF; s_array[j * 16 + 2] = (var.x >> 16) & 0x000000FF; s_array[j * 16 + 3] = (var.x >> 24) & 0x000000FF; My question is, is it possible to recast var (or for that matter *text4) to unsigned char in order to avoid the additional overhead of the bitwise operations?

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  • Given a vector of maximum 10 000 natural and distinct numbers, find 4 numbers(a, b, c, d) such that

    - by king_kong
    Hi, I solved this problem by following a straightforward but not optimal algorithm. I sorted the vector in descending order and after that substracted numbers from max to min to see if I get a + b + c = d. Notice that I haven't used anywhere the fact that elements are natural, distinct and 10 000 at most. I suppose these details are the key. Does anyone here have a hint over an optimal way of solving this? Thank you in advance! Later Edit: My idea goes like this: '<<quicksort in descending order>>' for i:=0 to count { // after sorting, loop through the array int d := v[i]; for j:=i+1 to count { int dif1 := d - v[j]; int a := v[j]; for k:=j+1 to count { if (v[k] > dif1) continue; int dif2 := dif1 - v[k]; b := v[k]; for l:=k+1 to count { if (dif2 = v[l]) { c := dif2; return {a, b, c, d} } } } } } What do you think?(sorry for the bad indentation)

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  • Isometric screen to 3D world coordinates efficiently

    - by Justin
    Been having a difficult time transforming 2D screen coordinates to 3D isometric space. This is the situation where I am working in 3D but I have an orthographic camera. Then my camera is positioned at (100, 200, 100), Where the xz plane is flat and y is up and down. I've been able to get a sort of working solution, but I feel like there must be a better way. Here's what I'm doing: With my camera at (0, 1, 0) I can translate my screen coordinates directly to 3D coordinates by doing: mouse2D.z = (( event.clientX / window.innerWidth ) * 2 - 1) * -(window.innerWidth /2); mouse2D.x = (( event.clientY / window.innerHeight) * 2 + 1) * -(window.innerHeight); mouse2D.y = 0; Everything okay so far. Now when I change my camera back to (100, 200, 100) my 3D space has been rotated 45 degrees around the y axis and then rotated about 54 degrees around a vector Q that runs along the xz plane at a 45 degree angle between the positive z axis and the negative x axis. So what I do to find the point is first rotate my point by 45 degrees using a matrix around the y axis. Now I'm close. So then I rotate my point around the vector Q. But my point is closer to the origin than it should be, since the Y value is not 0 anymore. What I want is that after the rotation my Y value is 0. So now I exchange my X and Z coordinates of my rotated vector with the X and Z coordinates of my non-rotated vector. So basically I have my old vector but it's y value is at an appropriate rotated amount. Now I use another matrix to rotate my point around the vector Q in the opposite direction, and I end up with the point where I clicked. Is there a better way? I feel like I must be missing something. Also my method isn't completely accurate. I feel like it's within 5-10 coordinates of where I click, maybe because of rounding from many calculations. Sorry for such a long question.

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  • Implement Negascout Algorithm with stack

    - by Dan
    I'm not familiar with how these stack exchange accounts work so if this is double posting I apologize. I asked the same thing on stackoverflow. I have added an AI routine to a game I am working on using the Negascout algorithm. It works great, but when I set a higher maximum depth it can take a few seconds to complete. The problem is it blocks the main thread, and the framework I am using does not have a way to deal with multi-threading properly across platforms. So I am trying to change this routine from recursively calling itself, to just managing a stack (vector) so that I can progress through the routine at a controlled pace and not lock up the application while the AI is thinking. I am getting hung up on the second recursive call in the loop. It relies on a returned value from the first call, so I don't know how to add these to a stack. My Working c++ Recursive Code: MoveScore abNegascout(vector<vector<char> > &board, int ply, int alpha, int beta, char piece) { if (ply==mMaxPly) { return MoveScore(evaluation.evaluateBoard(board, piece, oppPiece)); } int currentScore; int bestScore = -INFINITY; MoveCoord bestMove; int adaptiveBeta = beta; vector<MoveCoord> moveList = evaluation.genPriorityMoves(board, piece, findValidMove(board, piece, false)); if (moveList.empty()) { return MoveScore(bestScore); } bestMove = moveList[0]; for(int i=0;i<moveList.size();i++) { MoveCoord move = moveList[i]; vector<vector<char> > newBoard; newBoard.insert( newBoard.end(), board.begin(), board.end() ); effectMove(newBoard, piece, move.getRow(), move.getCol()); // First Call ****** MoveScore current = abNegascout(newBoard, ply+1, -adaptiveBeta, -max(alpha,bestScore), oppPiece); currentScore = - current.getScore(); if (currentScore>bestScore){ if (adaptiveBeta == beta || ply>=(mMaxPly-2)){ bestScore = currentScore; bestMove = move; }else { // Second Call ****** current = abNegascout(newBoard, ply+1, -beta, -currentScore, oppPiece); bestScore = - current.getScore(); bestMove = move; } if(bestScore>=beta){ return MoveScore(bestMove,bestScore); } adaptiveBeta = max(alpha, bestScore) + 1; } } return MoveScore(bestMove,bestScore); } If someone can please help by explaining how to get this to work with a simple stack. Example code would be much appreciated. While c++ would be perfect, any language that demonstrates how would be great. Thank You.

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  • GLSL: Strange light reflections [Solved]

    - by Tom
    According to this tutorial I'm trying to make a normal mapping using GLSL, but something is wrong and I can't find the solution. The output render is in this image: Image1 in this image is a plane with two triangles and each of it is different illuminated (that is bad). The plane has 6 vertices. In the upper left side of this plane are 2 identical vertices (same in the lower right). Here are some vectors same for each vertice: normal vector = 0, 1, 0 (red lines on image) tangent vector = 0, 0,-1 (green lines on image) bitangent vector = -1, 0, 0 (blue lines on image) here I have one question: The two identical vertices does need to have the same tangent and bitangent? I have tried to make other values to the tangents but the effect was still similar. Here are my shaders Vertex shader: #version 130 // Input vertex data, different for all executions of this shader. in vec3 vertexPosition_modelspace; in vec2 vertexUV; in vec3 vertexNormal_modelspace; in vec3 vertexTangent_modelspace; in vec3 vertexBitangent_modelspace; // Output data ; will be interpolated for each fragment. out vec2 UV; out vec3 Position_worldspace; out vec3 EyeDirection_cameraspace; out vec3 LightDirection_cameraspace; out vec3 LightDirection_tangentspace; out vec3 EyeDirection_tangentspace; // Values that stay constant for the whole mesh. uniform mat4 MVP; uniform mat4 V; uniform mat4 M; uniform mat3 MV3x3; uniform vec3 LightPosition_worldspace; void main(){ // Output position of the vertex, in clip space : MVP * position gl_Position = MVP * vec4(vertexPosition_modelspace,1); // Position of the vertex, in worldspace : M * position Position_worldspace = (M * vec4(vertexPosition_modelspace,1)).xyz; // Vector that goes from the vertex to the camera, in camera space. // In camera space, the camera is at the origin (0,0,0). vec3 vertexPosition_cameraspace = ( V * M * vec4(vertexPosition_modelspace,1)).xyz; EyeDirection_cameraspace = vec3(0,0,0) - vertexPosition_cameraspace; // Vector that goes from the vertex to the light, in camera space. M is ommited because it's identity. vec3 LightPosition_cameraspace = ( V * vec4(LightPosition_worldspace,1)).xyz; LightDirection_cameraspace = LightPosition_cameraspace + EyeDirection_cameraspace; // UV of the vertex. No special space for this one. UV = vertexUV; // model to camera = ModelView vec3 vertexTangent_cameraspace = MV3x3 * vertexTangent_modelspace; vec3 vertexBitangent_cameraspace = MV3x3 * vertexBitangent_modelspace; vec3 vertexNormal_cameraspace = MV3x3 * vertexNormal_modelspace; mat3 TBN = transpose(mat3( vertexTangent_cameraspace, vertexBitangent_cameraspace, vertexNormal_cameraspace )); // You can use dot products instead of building this matrix and transposing it. See References for details. LightDirection_tangentspace = TBN * LightDirection_cameraspace; EyeDirection_tangentspace = TBN * EyeDirection_cameraspace; } Fragment shader: #version 130 // Interpolated values from the vertex shaders in vec2 UV; in vec3 Position_worldspace; in vec3 EyeDirection_cameraspace; in vec3 LightDirection_cameraspace; in vec3 LightDirection_tangentspace; in vec3 EyeDirection_tangentspace; // Ouput data out vec3 color; // Values that stay constant for the whole mesh. uniform sampler2D DiffuseTextureSampler; uniform sampler2D NormalTextureSampler; uniform sampler2D SpecularTextureSampler; uniform mat4 V; uniform mat4 M; uniform mat3 MV3x3; uniform vec3 LightPosition_worldspace; void main(){ // Light emission properties // You probably want to put them as uniforms vec3 LightColor = vec3(1,1,1); float LightPower = 40.0; // Material properties vec3 MaterialDiffuseColor = texture2D( DiffuseTextureSampler, vec2(UV.x,-UV.y) ).rgb; vec3 MaterialAmbientColor = vec3(0.1,0.1,0.1) * MaterialDiffuseColor; //vec3 MaterialSpecularColor = texture2D( SpecularTextureSampler, UV ).rgb * 0.3; vec3 MaterialSpecularColor = vec3(0.5,0.5,0.5); // Local normal, in tangent space. V tex coordinate is inverted because normal map is in TGA (not in DDS) for better quality vec3 TextureNormal_tangentspace = normalize(texture2D( NormalTextureSampler, vec2(UV.x,-UV.y) ).rgb*2.0 - 1.0); // Distance to the light float distance = length( LightPosition_worldspace - Position_worldspace ); // Normal of the computed fragment, in camera space vec3 n = TextureNormal_tangentspace; // Direction of the light (from the fragment to the light) vec3 l = normalize(LightDirection_tangentspace); // Cosine of the angle between the normal and the light direction, // clamped above 0 // - light is at the vertical of the triangle -> 1 // - light is perpendicular to the triangle -> 0 // - light is behind the triangle -> 0 float cosTheta = clamp( dot( n,l ), 0,1 ); // Eye vector (towards the camera) vec3 E = normalize(EyeDirection_tangentspace); // Direction in which the triangle reflects the light vec3 R = reflect(-l,n); // Cosine of the angle between the Eye vector and the Reflect vector, // clamped to 0 // - Looking into the reflection -> 1 // - Looking elsewhere -> < 1 float cosAlpha = clamp( dot( E,R ), 0,1 ); color = // Ambient : simulates indirect lighting MaterialAmbientColor + // Diffuse : "color" of the object MaterialDiffuseColor * LightColor * LightPower * cosTheta / (distance*distance) + // Specular : reflective highlight, like a mirror MaterialSpecularColor * LightColor * LightPower * pow(cosAlpha,5) / (distance*distance); //color.xyz = E; //color.xyz = LightDirection_tangentspace; //color.xyz = EyeDirection_tangentspace; } I have replaced the original color value by EyeDirection_tangentspace vector and then I got other strange effect but I can not link the image (not eunogh reputation) Is it possible that with this shaders is something wrong, or maybe in other place in my code e.g with my matrices?

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  • how to reset monitor display settings

    - by vector
    On Ubuntu 12.4 laptop with Gnome desktop, I tried to set an additional display through Catalyst (administrative) and slowly ended up making making a mess out of the whole thing. I tried several combinations of settings and at each iterration I just made things worse, ending up with 'mail battery sound time user power' icons repeating on the top bar. Now I'm lost as to how to restore everything to default settings.

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