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• Image SEO - always repeat main keyword in alt text?

  - by Marcus Edensky

  I'm working on an Easter Island website and I'm currently redesigning my image system. Virtually all my photos are of Easter Island. My question is, should I always include the keywords "Easter Island" for Google to easier understand that my photos are from Easter Island, or is it sufficient that the "Easter Island" keywords are in the domain, as well as in all other pages of the site? For example, Alt text 1: "Moai statues at volcano Rano Raraku at Easter Island (Rapa Nui)" or Alt text 2: "Moai statues at volcano Rano Raraku" Would example 1 be considered keyword stuffing by Google

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• Java code optimization on matrix windowing computes in more time

  - by rano

  I have a matrix which represents an image and I need to cycle over each pixel and for each one of those I have to compute the sum of all its neighbors, ie the pixels that belong to a window of radius rad centered on the pixel. I came up with three alternatives: The simplest way, the one that recomputes the window for each pixel The more optimized way that uses a queue to store the sums of the window columns and cycling through the columns of the matrix updates this queue by adding a new element and removing the oldes The even more optimized way that does not need to recompute the queue for each row but incrementally adjusts a previously saved one I implemented them in c++ using a queue for the second method and a combination of deques for the third (I need to iterate through their elements without destructing them) and scored their times to see if there was an actual improvement. it appears that the third method is indeed faster. Then I tried to port the code to Java (and I must admit that I'm not very comfortable with it). I used ArrayDeque for the second method and LinkedLists for the third resulting in the third being inefficient in time. Here is the simplest method in C++ (I'm not posting the java version since it is almost identical): void normalWindowing(int mat[][MAX], int cols, int rows, int rad){ int i, j; int h = 0; for (i = 0; i < rows; ++i) { for (j = 0; j < cols; j++) { h = 0; for (int ry =- rad; ry <= rad; ry++) { int y = i + ry; if (y >= 0 && y < rows) { for (int rx =- rad; rx <= rad; rx++) { int x = j + rx; if (x >= 0 && x < cols) { h += mat[y][x]; } } } } } } } Here is the second method (the one optimized through columns) in C++: void opt1Windowing(int mat[][MAX], int cols, int rows, int rad){ int i, j, h, y, col; queue<int>* q = NULL; for (i = 0; i < rows; ++i) { if (q != NULL) delete(q); q = new queue<int>(); h = 0; for (int rx = 0; rx <= rad; rx++) { if (rx < cols) { int mem = 0; for (int ry =- rad; ry <= rad; ry++) { y = i + ry; if (y >= 0 && y < rows) { mem += mat[y][rx]; } } q->push(mem); h += mem; } } for (j = 1; j < cols; j++) { col = j + rad; if (j - rad > 0) { h -= q->front(); q->pop(); } if (j + rad < cols) { int mem = 0; for (int ry =- rad; ry <= rad; ry++) { y = i + ry; if (y >= 0 && y < rows) { mem += mat[y][col]; } } q->push(mem); h += mem; } } } } And here is the Java version: public static void opt1Windowing(int [][] mat, int rad){ int i, j = 0, h, y, col; int cols = mat[0].length; int rows = mat.length; ArrayDeque<Integer> q = null; for (i = 0; i < rows; ++i) { q = new ArrayDeque<Integer>(); h = 0; for (int rx = 0; rx <= rad; rx++) { if (rx < cols) { int mem = 0; for (int ry =- rad; ry <= rad; ry++) { y = i + ry; if (y >= 0 && y < rows) { mem += mat[y][rx]; } } q.addLast(mem); h += mem; } } j = 0; for (j = 1; j < cols; j++) { col = j + rad; if (j - rad > 0) { h -= q.peekFirst(); q.pop(); } if (j + rad < cols) { int mem = 0; for (int ry =- rad; ry <= rad; ry++) { y = i + ry; if (y >= 0 && y < rows) { mem += mat[y][col]; } } q.addLast(mem); h += mem; } } } } I recognize this post will be a wall of text. Here is the third method in C++: void opt2Windowing(int mat[][MAX], int cols, int rows, int rad){ int i = 0; int j = 0; int h = 0; int hh = 0; deque< deque<int> *> * M = new deque< deque<int> *>(); for (int ry = 0; ry <= rad; ry++) { if (ry < rows) { deque<int> * q = new deque<int>(); M->push_back(q); for (int rx = 0; rx <= rad; rx++) { if (rx < cols) { int val = mat[ry][rx]; q->push_back(val); h += val; } } } } deque<int> * C = new deque<int>(M->front()->size()); deque<int> * Q = new deque<int>(M->front()->size()); deque<int> * R = new deque<int>(M->size()); deque< deque<int> *>::iterator mit; deque< deque<int> *>::iterator mstart = M->begin(); deque< deque<int> *>::iterator mend = M->end(); deque<int>::iterator rit; deque<int>::iterator rstart = R->begin(); deque<int>::iterator rend = R->end(); deque<int>::iterator cit; deque<int>::iterator cstart = C->begin(); deque<int>::iterator cend = C->end(); for (mit = mstart, rit = rstart; mit != mend, rit != rend; ++mit, ++rit) { deque<int>::iterator pit; deque<int>::iterator pstart = (* mit)->begin(); deque<int>::iterator pend = (* mit)->end(); for(cit = cstart, pit = pstart; cit != cend && pit != pend; ++cit, ++pit) { (* cit) += (* pit); (* rit) += (* pit); } } for (i = 0; i < rows; ++i) { j = 0; if (i - rad > 0) { deque<int>::iterator cit; deque<int>::iterator cstart = C->begin(); deque<int>::iterator cend = C->end(); deque<int>::iterator pit; deque<int>::iterator pstart = (M->front())->begin(); deque<int>::iterator pend = (M->front())->end(); for(cit = cstart, pit = pstart; cit != cend; ++cit, ++pit) { (* cit) -= (* pit); } deque<int> * k = M->front(); M->pop_front(); delete k; h -= R->front(); R->pop_front(); } int row = i + rad; if (row < rows && i > 0) { deque<int> * newQ = new deque<int>(); M->push_back(newQ); deque<int>::iterator cit; deque<int>::iterator cstart = C->begin(); deque<int>::iterator cend = C->end(); int rx; int tot = 0; for (rx = 0, cit = cstart; rx <= rad; rx++, ++cit) { if (rx < cols) { int val = mat[row][rx]; newQ->push_back(val); (* cit) += val; tot += val; } } R->push_back(tot); h += tot; } hh = h; copy(C->begin(), C->end(), Q->begin()); for (j = 1; j < cols; j++) { int col = j + rad; if (j - rad > 0) { hh -= Q->front(); Q->pop_front(); } if (j + rad < cols) { int val = 0; for (int ry =- rad; ry <= rad; ry++) { int y = i + ry; if (y >= 0 && y < rows) { val += mat[y][col]; } } hh += val; Q->push_back(val); } } } } And finally its Java version: public static void opt2Windowing(int [][] mat, int rad){ int cols = mat[0].length; int rows = mat.length; int i = 0; int j = 0; int h = 0; int hh = 0; LinkedList<LinkedList<Integer>> M = new LinkedList<LinkedList<Integer>>(); for (int ry = 0; ry <= rad; ry++) { if (ry < rows) { LinkedList<Integer> q = new LinkedList<Integer>(); M.addLast(q); for (int rx = 0; rx <= rad; rx++) { if (rx < cols) { int val = mat[ry][rx]; q.addLast(val); h += val; } } } } int firstSize = M.getFirst().size(); int mSize = M.size(); LinkedList<Integer> C = new LinkedList<Integer>(); LinkedList<Integer> Q = null; LinkedList<Integer> R = new LinkedList<Integer>(); for (int k = 0; k < firstSize; k++) { C.add(0); } for (int k = 0; k < mSize; k++) { R.add(0); } ListIterator<LinkedList<Integer>> mit; ListIterator<Integer> rit; ListIterator<Integer> cit; ListIterator<Integer> pit; for (mit = M.listIterator(), rit = R.listIterator(); mit.hasNext();) { Integer r = rit.next(); int rsum = 0; for (cit = C.listIterator(), pit = (mit.next()).listIterator(); cit.hasNext();) { Integer c = cit.next(); Integer p = pit.next(); rsum += p; cit.set(c + p); } rit.set(r + rsum); } for (i = 0; i < rows; ++i) { j = 0; if (i - rad > 0) { for(cit = C.listIterator(), pit = M.getFirst().listIterator(); cit.hasNext();) { Integer c = cit.next(); Integer p = pit.next(); cit.set(c - p); } M.removeFirst(); h -= R.getFirst(); R.removeFirst(); } int row = i + rad; if (row < rows && i > 0) { LinkedList<Integer> newQ = new LinkedList<Integer>(); M.addLast(newQ); int rx; int tot = 0; for (rx = 0, cit = C.listIterator(); rx <= rad; rx++) { if (rx < cols) { Integer c = cit.next(); int val = mat[row][rx]; newQ.addLast(val); cit.set(c + val); tot += val; } } R.addLast(tot); h += tot; } hh = h; Q = new LinkedList<Integer>(); Q.addAll(C); for (j = 1; j < cols; j++) { int col = j + rad; if (j - rad > 0) { hh -= Q.getFirst(); Q.pop(); } if (j + rad < cols) { int val = 0; for (int ry =- rad; ry <= rad; ry++) { int y = i + ry; if (y >= 0 && y < rows) { val += mat[y][col]; } } hh += val; Q.addLast(val); } } } } I guess that most is due to the poor choice of the LinkedList in Java and to the lack of an efficient (not shallow) copy method between two LinkedList. How can I improve the third Java method? Am I doing some conceptual error? As always, any criticisms is welcome. UPDATE Even if it does not solve the issue, using ArrayLists, as being suggested, instead of LinkedList improves the third method. The second one performs still better (but when the number of rows and columns of the matrix is lower than 300 and the window radius is small the first unoptimized method is the fastest in Java)

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• Java code optimization leads to numerical inaccuracies and errors

  - by rano

  I'm trying to implement a version of the Fuzzy C-Means algorithm in Java and I'm trying to do some optimization by computing just once everything that can be computed just once. This is an iterative algorithm and regarding the updating of a matrix, the clusters x pixels membership matrix U, this is the update rule I want to optimize: where the x are the element of a matrix X (pixels x features) and v belongs to the matrix V (clusters x features). And m is a parameter that ranges from 1.1 to infinity. The distance used is the euclidean norm. If I had to implement this formula in a banal way I'd do: for(int i = 0; i < X.length; i++) { int count = 0; for(int j = 0; j < V.length; j++) { double num = D[i][j]; double sumTerms = 0; for(int k = 0; k < V.length; k++) { double thisDistance = D[i][k]; sumTerms += Math.pow(num / thisDistance, (1.0 / (m - 1.0))); } U[i][j] = (float) (1f / sumTerms); } } In this way some optimization is already done, I precomputed all the possible squared distances between X and V and stored them in a matrix D but that is not enough, since I'm cycling througn the elements of V two times resulting in two nested loops. Looking at the formula the numerator of the fraction is independent of the sum so I can compute numerator and denominator independently and the denominator can be computed just once for each pixel. So I came to a solution like this: int nClusters = V.length; double exp = (1.0 / (m - 1.0)); for(int i = 0; i < X.length; i++) { int count = 0; for(int j = 0; j < nClusters; j++) { double distance = D[i][j]; double denominator = D[i][nClusters]; double numerator = Math.pow(distance, exp); U[i][j] = (float) (1f / (numerator * denominator)); } } Where I precomputed the denominator into an additional column of the matrix D while I was computing the distances: for (int i = 0; i < X.length; i++) { for (int j = 0; j < V.length; j++) { double sum = 0; for (int k = 0; k < nDims; k++) { final double d = X[i][k] - V[j][k]; sum += d * d; } D[i][j] = sum; D[i][B.length] += Math.pow(1 / D[i][j], exp); } } By doing so I encounter numerical differences between the 'banal' computation and the second one that leads to different numerical value in U (not in the first iterates but soon enough). I guess that the problem is that exponentiate very small numbers to high values (the elements of U can range from 0.0 to 1.0 and exp , for m = 1.1, is 10) leads to ver y small values, whereas by dividing the numerator and the denominator and THEN exponentiating the result seems to be better numerically. The problem is it involves much more operations. Am I doing something wrong? Is there a possible solution to get both the code optimized and numerically stable? Any suggestion or criticism will be appreciated.

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• How To Remove People and Objects From Photographs In Photoshop

  - by Eric Z Goodnight

  You might think that it’s a complicated process to remove objects from photographs. But really Photoshop makes it quite simple, even when removing all traces of a person from digital photographs. Read on to see just how easy it is. Photoshop was originally created to be an image editing program, and it excels at it. With hardly any Photoshop experience, any beginner can begin removing objects or people from their photos. Have some friends that photobombed an otherwise great pic? Tell them to say their farewells, because here’s how to get rid of them with Photoshop! Tools for Removing Objects Removing an object is not really “magical” work. Your goal is basically to cover up the information you don’t want in an image with information you do want. In this sample image, we want to remove the cigar smoking man, and leave the geisha. Here’s a couple of the tools that can be useful to work with when attempting this kind of task. Clone Stamp and Pattern Stamp Tool: Samples parts of your image from your background, and allows you to paint into your image with your mouse or stylus. Eraser and Brush Tools: Paint flat colors and shapes, and erase cloned layers of image information. Basic, down and dirty photo editing tools. Pen, Quick Selection, Lasso, and Crop tools: Select, isolate, and remove parts of your image with these selection tools. All useful in their own way. Some, like the pen tool, are nightmarishly tough on beginners. Remove a Person with the Clone Stamp Tool (Video) The video above uses the Clone Stamp tool to sample and paint with the background texture. It’s a simple tool to use, although it can be confusing, possibly counter-intuitive. Here’s some pointers, in addition to the video above. Select shortcut key to choose the Clone tool stamp from the Tools Panel. Always create a copy of your background layer before doing heavy edits by right clicking on the background in your Layers Panel and selecting “Duplicate.” Hold with the Clone Tool selected, and click anywhere in your image to sample that area. When you’re sampling an area, your cursor is “Aligned” with your sample area. When you paint, your sample area moves. You can turn the “Aligned” setting off by clicking the in the Options Panel at the top of your screen if you want. Change your brush size and hardness as shown in the video by right-clicking in your image. Use your lasso to copy and paste pieces of your image in order to cover up any parts that seem appropriate. Photoshop Magic with the “Content-Aware Fill” One of the hallmark features of CS5 is the “Content-Aware Fill.” Content aware fill can be an excellent shortcut to removing objects and even people in Photoshop, but it is somewhat limited, and can get confused. Here’s a basic rundown on how it works. Select an object using your Lasso tool, shortcut key . The Lasso works fine as this selection can be rough. Navigate to Edit > Fill, and select “Content-Aware,” as illustrated above, from the pull-down menu. It’s surprisingly simple. After some processing, Photoshop has done the work of removing the object for you. It takes a few moments, and it is not perfect, so be prepared to touch it up with some Copy-Paste, or some Clone stamp action. Content Aware Fill Has Its Limits Keep in mind that the Content Aware Fill is meant to be used with other techniques in mind. It doesn’t always perform perfectly, but can give you a great starting point. Take this image for instance. It is actually plausible to hide this figure and make this image look like he was never there at all. With a selection made with the Lasso tool, navigate to Edit > Fill and select “Content Aware” again. The result is surprisingly good, but as you can see, worthy of some touch up. With a result like this one, you’ll have to get your hands dirty with copy-paste to create believable lines in the background. With many photographs, Content Aware Fill will simply get confused and give you results you won’t be happy with. Additional Touch Up for Bad Background Textures with the Pattern Stamp Tool For the perfectionist, cleaning up the lumpy looking textures that the Clone Stamp can leave is fairly simple using the Pattern Stamp Tool. Sample an piece of your image with your Marquee Tool, shortcut key . Navigate to Edit > Define Pattern to create a new Pattern from your selection. Click OK to continue. Click and hold down on the Clone Stamp tool in your Tools Panel until you can select the Pattern Stamp Tool. Pick your new pattern from the Options at the top of your screen, in the Options Panel. Then simply right click in your image in order to pick as soft a brush as possible to paint with. Paint into your image until your background is as smooth as you want it to be, making your painted out object more and more invisible. If you get lines from your repeated texture, experiment turning the on and off and paint over them. In addition to this, simple use of the Crop Tool, shortcut , can recompose an image, making it look as if it never had another object in it at all. Combine these techniques to find a method that works best for your images. Have questions or comments concerning Graphics, Photos, Filetypes, or Photoshop? Send your questions to [email protected], and they may be featured in a future How-To Geek Graphics article. Image Credits: Geisha Kyoto Gion by Todd Laracuenta via Wikipedia, used under Creative Commons. Moai Rano raraku by Aurbina, in Public Domain. Chris Young visits Wrigley by TonyTheTiger, via Wikipedia, used under Creative Commons. Latest Features How-To Geek ETC Ask How-To Geek: How Can I Monitor My Bandwidth Usage? 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