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  • The Incremental Architect&acute;s Napkin &ndash; #3 &ndash; Make Evolvability inevitable

    - by Ralf Westphal
    Originally posted on: http://geekswithblogs.net/theArchitectsNapkin/archive/2014/06/04/the-incremental-architectacutes-napkin-ndash-3-ndash-make-evolvability-inevitable.aspxThe easier something to measure the more likely it will be produced. Deviations between what is and what should be can be readily detected. That´s what automated acceptance tests are for. That´s what sprint reviews in Scrum are for. It´s no small wonder our software looks like it looks. It has all the traits whose conformance with requirements can easily be measured. And it´s lacking traits which cannot easily be measured. Evolvability (or Changeability) is such a trait. If an operation is correct, if an operation if fast enough, that can be checked very easily. But whether Evolvability is high or low, that cannot be checked by taking a measure or two. Evolvability might correlate with certain traits, e.g. number of lines of code (LOC) per function or Cyclomatic Complexity or test coverage. But there is no threshold value signalling “evolvability too low”; also Evolvability is hardly tangible for the customer. Nevertheless Evolvability is of great importance - at least in the long run. You can get away without much of it for a short time. Eventually, though, it´s needed like any other requirement. Or even more. Because without Evolvability no other requirement can be implemented. Evolvability is the foundation on which all else is build. Such fundamental importance is in stark contrast with its immeasurability. To compensate this, Evolvability must be put at the very center of software development. It must become the hub around everything else revolves. Since we cannot measure Evolvability, though, we cannot start watching it more. Instead we need to establish practices to keep it high (enough) at all times. Chefs have known that for long. That´s why everybody in a restaurant kitchen is constantly seeing after cleanliness. Hygiene is important as is to have clean tools at standardized locations. Only then the health of the patrons can be guaranteed and production efficiency is constantly high. Still a kitchen´s level of cleanliness is easier to measure than software Evolvability. That´s why important practices like reviews, pair programming, or TDD are not enough, I guess. What we need to keep Evolvability in focus and high is… to continually evolve. Change must not be something to avoid but too embrace. To me that means the whole change cycle from requirement analysis to delivery needs to be gone through more often. Scrum´s sprints of 4, 2 even 1 week are too long. Kanban´s flow of user stories across is too unreliable; it takes as long as it takes. Instead we should fix the cycle time at 2 days max. I call that Spinning. No increment must take longer than from this morning until tomorrow evening to finish. Then it should be acceptance checked by the customer (or his/her representative, e.g. a Product Owner). For me there are several resasons for such a fixed and short cycle time for each increment: Clear expectations Absolute estimates (“This will take X days to complete.”) are near impossible in software development as explained previously. Too much unplanned research and engineering work lurk in every feature. And then pervasive interruptions of work by peers and management. However, the smaller the scope the better our absolute estimates become. That´s because we understand better what really are the requirements and what the solution should look like. But maybe more importantly the shorter the timespan the more we can control how we use our time. So much can happen over the course of a week and longer timespans. But if push comes to shove I can block out all distractions and interruptions for a day or possibly two. That´s why I believe we can give rough absolute estimates on 3 levels: Noon Tonight Tomorrow Think of a meeting with a Product Owner at 8:30 in the morning. If she asks you, how long it will take you to implement a user story or bug fix, you can say, “It´ll be fixed by noon.”, or you can say, “I can manage to implement it until tonight before I leave.”, or you can say, “You´ll get it by tomorrow night at latest.” Yes, I believe all else would be naive. If you´re not confident to get something done by tomorrow night (some 34h from now) you just cannot reliably commit to any timeframe. That means you should not promise anything, you should not even start working on the issue. So when estimating use these four categories: Noon, Tonight, Tomorrow, NoClue - with NoClue meaning the requirement needs to be broken down further so each aspect can be assigned to one of the first three categories. If you like absolute estimates, here you go. But don´t do deep estimates. Don´t estimate dozens of issues; don´t think ahead (“Issue A is a Tonight, then B will be a Tomorrow, after that it´s C as a Noon, finally D is a Tonight - that´s what I´ll do this week.”). Just estimate so Work-in-Progress (WIP) is 1 for everybody - plus a small number of buffer issues. To be blunt: Yes, this makes promises impossible as to what a team will deliver in terms of scope at a certain date in the future. But it will give a Product Owner a clear picture of what to pull for acceptance feedback tonight and tomorrow. Trust through reliability Our trade is lacking trust. Customers don´t trust software companies/departments much. Managers don´t trust developers much. I find that perfectly understandable in the light of what we´re trying to accomplish: delivering software in the face of uncertainty by means of material good production. Customers as well as managers still expect software development to be close to production of houses or cars. But that´s a fundamental misunderstanding. Software development ist development. It´s basically research. As software developers we´re constantly executing experiments to find out what really provides value to users. We don´t know what they need, we just have mediated hypothesises. That´s why we cannot reliably deliver on preposterous demands. So trust is out of the window in no time. If we switch to delivering in short cycles, though, we can regain trust. Because estimates - explicit or implicit - up to 32 hours at most can be satisfied. I´d say: reliability over scope. It´s more important to reliably deliver what was promised then to cover a lot of requirement area. So when in doubt promise less - but deliver without delay. Deliver on scope (Functionality and Quality); but also deliver on Evolvability, i.e. on inner quality according to accepted principles. Always. Trust will be the reward. Less complexity of communication will follow. More goodwill buffer will follow. So don´t wait for some Kanban board to show you, that flow can be improved by scheduling smaller stories. You don´t need to learn that the hard way. Just start with small batch sizes of three different sizes. Fast feedback What has been finished can be checked for acceptance. Why wait for a sprint of several weeks to end? Why let the mental model of the issue and its solution dissipate? If you get final feedback after one or two weeks, you hardly remember what you did and why you did it. Resoning becomes hard. But more importantly youo probably are not in the mood anymore to go back to something you deemed done a long time ago. It´s boring, it´s frustrating to open up that mental box again. Learning is harder the longer it takes from event to feedback. Effort can be wasted between event (finishing an issue) and feedback, because other work might go in the wrong direction based on false premises. Checking finished issues for acceptance is the most important task of a Product Owner. It´s even more important than planning new issues. Because as long as work started is not released (accepted) it´s potential waste. So before starting new work better make sure work already done has value. By putting the emphasis on acceptance rather than planning true pull is established. As long as planning and starting work is more important, it´s a push process. Accept a Noon issue on the same day before leaving. Accept a Tonight issue before leaving today or first thing tomorrow morning. Accept a Tomorrow issue tomorrow night before leaving or early the day after tomorrow. After acceptance the developer(s) can start working on the next issue. Flexibility As if reliability/trust and fast feedback for less waste weren´t enough economic incentive, there is flexibility. After each issue the Product Owner can change course. If on Monday morning feature slices A, B, C, D, E were important and A, B, C were scheduled for acceptance by Monday evening and Tuesday evening, the Product Owner can change her mind at any time. Maybe after A got accepted she asks for continuation with D. But maybe, just maybe, she has gotten a completely different idea by then. Maybe she wants work to continue on F. And after B it´s neither D nor E, but G. And after G it´s D. With Spinning every 32 hours at latest priorities can be changed. And nothing is lost. Because what got accepted is of value. It provides an incremental value to the customer/user. Or it provides internal value to the Product Owner as increased knowledge/decreased uncertainty. I find such reactivity over commitment economically very benefical. Why commit a team to some workload for several weeks? It´s unnecessary at beast, and inflexible and wasteful at worst. If we cannot promise delivery of a certain scope on a certain date - which is what customers/management usually want -, we can at least provide them with unpredecented flexibility in the face of high uncertainty. Where the path is not clear, cannot be clear, make small steps so you´re able to change your course at any time. Premature completion Customers/management are used to premeditating budgets. They want to know exactly how much to pay for a certain amount of requirements. That´s understandable. But it does not match with the nature of software development. We should know that by now. Maybe there´s somewhere in the world some team who can consistently deliver on scope, quality, and time, and budget. Great! Congratulations! I, however, haven´t seen such a team yet. Which does not mean it´s impossible, but I think it´s nothing I can recommend to strive for. Rather I´d say: Don´t try this at home. It might hurt you one way or the other. However, what we can do, is allow customers/management stop work on features at any moment. With spinning every 32 hours a feature can be declared as finished - even though it might not be completed according to initial definition. I think, progress over completion is an important offer software development can make. Why think in terms of completion beyond a promise for the next 32 hours? Isn´t it more important to constantly move forward? Step by step. We´re not running sprints, we´re not running marathons, not even ultra-marathons. We´re in the sport of running forever. That makes it futile to stare at the finishing line. The very concept of a burn-down chart is misleading (in most cases). Whoever can only think in terms of completed requirements shuts out the chance for saving money. The requirements for a features mostly are uncertain. So how does a Product Owner know in the first place, how much is needed. Maybe more than specified is needed - which gets uncovered step by step with each finished increment. Maybe less than specified is needed. After each 4–32 hour increment the Product Owner can do an experient (or invite users to an experiment) if a particular trait of the software system is already good enough. And if so, she can switch the attention to a different aspect. In the end, requirements A, B, C then could be finished just 70%, 80%, and 50%. What the heck? It´s good enough - for now. 33% money saved. Wouldn´t that be splendid? Isn´t that a stunning argument for any budget-sensitive customer? You can save money and still get what you need? Pull on practices So far, in addition to more trust, more flexibility, less money spent, Spinning led to “doing less” which also means less code which of course means higher Evolvability per se. Last but not least, though, I think Spinning´s short acceptance cycles have one more effect. They excert pull-power on all sorts of practices known for increasing Evolvability. If, for example, you believe high automated test coverage helps Evolvability by lowering the fear of inadverted damage to a code base, why isn´t 90% of the developer community practicing automated tests consistently? I think, the answer is simple: Because they can do without. Somehow they manage to do enough manual checks before their rare releases/acceptance checks to ensure good enough correctness - at least in the short term. The same goes for other practices like component orientation, continuous build/integration, code reviews etc. None of that is compelling, urgent, imperative. Something else always seems more important. So Evolvability principles and practices fall through the cracks most of the time - until a project hits a wall. Then everybody becomes desperate; but by then (re)gaining Evolvability has become as very, very difficult and tedious undertaking. Sometimes up to the point where the existence of a project/company is in danger. With Spinning that´s different. If you´re practicing Spinning you cannot avoid all those practices. With Spinning you very quickly realize you cannot deliver reliably even on your 32 hour promises. Spinning thus is pulling on developers to adopt principles and practices for Evolvability. They will start actively looking for ways to keep their delivery rate high. And if not, management will soon tell them to do that. Because first the Product Owner then management will notice an increasing difficulty to deliver value within 32 hours. There, finally there emerges a way to measure Evolvability: The more frequent developers tell the Product Owner there is no way to deliver anything worth of feedback until tomorrow night, the poorer Evolvability is. Don´t count the “WTF!”, count the “No way!” utterances. In closing For sustainable software development we need to put Evolvability first. Functionality and Quality must not rule software development but be implemented within a framework ensuring (enough) Evolvability. Since Evolvability cannot be measured easily, I think we need to put software development “under pressure”. Software needs to be changed more often, in smaller increments. Each increment being relevant to the customer/user in some way. That does not mean each increment is worthy of shipment. It´s sufficient to gain further insight from it. Increments primarily serve the reduction of uncertainty, not sales. Sales even needs to be decoupled from this incremental progress. No more promises to sales. No more delivery au point. Rather sales should look at a stream of accepted increments (or incremental releases) and scoup from that whatever they find valuable. Sales and marketing need to realize they should work on what´s there, not what might be possible in the future. But I digress… In my view a Spinning cycle - which is not easy to reach, which requires practice - is the core practice to compensate the immeasurability of Evolvability. From start to finish of each issue in 32 hours max - that´s the challenge we need to accept if we´re serious increasing Evolvability. Fortunately higher Evolvability is not the only outcome of Spinning. Customer/management will like the increased flexibility and “getting more bang for the buck”.

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  • Finding the normal of OBB face with an OBB penetrating

    - by Milo
    Below is an illustration: I have an OBB in an OBB (see below for OBB2D code if needed). What I need to determine is, what face it is in, and what direction do I point the normal? The goal is to get the OBB out of the OBB so the normal needs to face outward of the OBB. How could I go about: Finding what face the line is penetrating given the 4 corners of the OBB and the class below: if we define dx=x2-x1 and dy=y2-y1, then the normals are (-dy, dx) and (dy, -dx). Which normal points outward of the OBB? Thanks public class OBB2D { // Corners of the box, where 0 is the lower left. private Vector2D corner[] = new Vector2D[4]; private Vector2D center = new Vector2D(); private Vector2D extents = new Vector2D(); private RectF boundingRect = new RectF(); private float angle; //Two edges of the box extended away from corner[0]. private Vector2D axis[] = new Vector2D[2]; private double origin[] = new double[2]; public OBB2D(Vector2D center, float w, float h, float angle) { set(center,w,h,angle); } public OBB2D(float left, float top, float width, float height) { set(new Vector2D(left + (width / 2), top + (height / 2)),width,height,0.0f); } public void set(Vector2D center,float w, float h,float angle) { Vector2D X = new Vector2D( (float)Math.cos(angle), (float)Math.sin(angle)); Vector2D Y = new Vector2D((float)-Math.sin(angle), (float)Math.cos(angle)); X = X.multiply( w / 2); Y = Y.multiply( h / 2); corner[0] = center.subtract(X).subtract(Y); corner[1] = center.add(X).subtract(Y); corner[2] = center.add(X).add(Y); corner[3] = center.subtract(X).add(Y); computeAxes(); extents.x = w / 2; extents.y = h / 2; computeDimensions(center,angle); } private void computeDimensions(Vector2D center,float angle) { this.center.x = center.x; this.center.y = center.y; this.angle = angle; boundingRect.left = Math.min(Math.min(corner[0].x, corner[3].x), Math.min(corner[1].x, corner[2].x)); boundingRect.top = Math.min(Math.min(corner[0].y, corner[1].y),Math.min(corner[2].y, corner[3].y)); boundingRect.right = Math.max(Math.max(corner[1].x, corner[2].x), Math.max(corner[0].x, corner[3].x)); boundingRect.bottom = Math.max(Math.max(corner[2].y, corner[3].y),Math.max(corner[0].y, corner[1].y)); } public void set(RectF rect) { set(new Vector2D(rect.centerX(),rect.centerY()),rect.width(),rect.height(),0.0f); } // Returns true if other overlaps one dimension of this. private boolean overlaps1Way(OBB2D other) { for (int a = 0; a < axis.length; ++a) { double t = other.corner[0].dot(axis[a]); // Find the extent of box 2 on axis a double tMin = t; double tMax = t; for (int c = 1; c < corner.length; ++c) { t = other.corner[c].dot(axis[a]); if (t < tMin) { tMin = t; } else if (t > tMax) { tMax = t; } } // We have to subtract off the origin // See if [tMin, tMax] intersects [0, 1] if ((tMin > 1 + origin[a]) || (tMax < origin[a])) { // There was no intersection along this dimension; // the boxes cannot possibly overlap. return false; } } // There was no dimension along which there is no intersection. // Therefore the boxes overlap. return true; } //Updates the axes after the corners move. Assumes the //corners actually form a rectangle. private void computeAxes() { axis[0] = corner[1].subtract(corner[0]); axis[1] = corner[3].subtract(corner[0]); // Make the length of each axis 1/edge length so we know any // dot product must be less than 1 to fall within the edge. for (int a = 0; a < axis.length; ++a) { axis[a] = axis[a].divide((axis[a].length() * axis[a].length())); origin[a] = corner[0].dot(axis[a]); } } public void moveTo(Vector2D center) { Vector2D centroid = (corner[0].add(corner[1]).add(corner[2]).add(corner[3])).divide(4.0f); Vector2D translation = center.subtract(centroid); for (int c = 0; c < 4; ++c) { corner[c] = corner[c].add(translation); } computeAxes(); computeDimensions(center,angle); } // Returns true if the intersection of the boxes is non-empty. public boolean overlaps(OBB2D other) { if(right() < other.left()) { return false; } if(bottom() < other.top()) { return false; } if(left() > other.right()) { return false; } if(top() > other.bottom()) { return false; } if(other.getAngle() == 0.0f && getAngle() == 0.0f) { return true; } return overlaps1Way(other) && other.overlaps1Way(this); } public Vector2D getCenter() { return center; } public float getWidth() { return extents.x * 2; } public float getHeight() { return extents.y * 2; } public void setAngle(float angle) { set(center,getWidth(),getHeight(),angle); } public float getAngle() { return angle; } public void setSize(float w,float h) { set(center,w,h,angle); } public float left() { return boundingRect.left; } public float right() { return boundingRect.right; } public float bottom() { return boundingRect.bottom; } public float top() { return boundingRect.top; } public RectF getBoundingRect() { return boundingRect; } public boolean overlaps(float left, float top, float right, float bottom) { if(right() < left) { return false; } if(bottom() < top) { return false; } if(left() > right) { return false; } if(top() > bottom) { return false; } return true; } };

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  • Move penetrating OBB out of another OBB to resolve collision

    - by Milo
    I'm working on collision resolution for my game. I just need a good way to get an object out of another object if it gets stuck. In this case a car. Here is a typical scenario. The red car is in the green object. How do I correctly get it out so the car can slide along the edge of the object as it should. I tried: if(buildings.size() > 0) { Entity e = buildings.get(0); Vector2D vel = new Vector2D(); vel.x = vehicle.getVelocity().x; vel.y = vehicle.getVelocity().y; vel.normalize(); while(vehicle.getRect().overlaps(e.getRect())) { vehicle.setCenter(vehicle.getCenterX() - vel.x * 0.1f, vehicle.getCenterY() - vel.y * 0.1f); } colided = true; } But that does not work too well. Is there some sort of vector I could calculate to use as the vector to move the car away from the object? Thanks Here is my OBB2D class: public class OBB2D { // Corners of the box, where 0 is the lower left. private Vector2D corner[] = new Vector2D[4]; private Vector2D center = new Vector2D(); private Vector2D extents = new Vector2D(); private RectF boundingRect = new RectF(); private float angle; //Two edges of the box extended away from corner[0]. private Vector2D axis[] = new Vector2D[2]; private double origin[] = new double[2]; public OBB2D(Vector2D center, float w, float h, float angle) { set(center,w,h,angle); } public OBB2D(float left, float top, float width, float height) { set(new Vector2D(left + (width / 2), top + (height / 2)),width,height,0.0f); } public void set(Vector2D center,float w, float h,float angle) { Vector2D X = new Vector2D( (float)Math.cos(angle), (float)Math.sin(angle)); Vector2D Y = new Vector2D((float)-Math.sin(angle), (float)Math.cos(angle)); X = X.multiply( w / 2); Y = Y.multiply( h / 2); corner[0] = center.subtract(X).subtract(Y); corner[1] = center.add(X).subtract(Y); corner[2] = center.add(X).add(Y); corner[3] = center.subtract(X).add(Y); computeAxes(); extents.x = w / 2; extents.y = h / 2; computeDimensions(center,angle); } private void computeDimensions(Vector2D center,float angle) { this.center.x = center.x; this.center.y = center.y; this.angle = angle; boundingRect.left = Math.min(Math.min(corner[0].x, corner[3].x), Math.min(corner[1].x, corner[2].x)); boundingRect.top = Math.min(Math.min(corner[0].y, corner[1].y),Math.min(corner[2].y, corner[3].y)); boundingRect.right = Math.max(Math.max(corner[1].x, corner[2].x), Math.max(corner[0].x, corner[3].x)); boundingRect.bottom = Math.max(Math.max(corner[2].y, corner[3].y),Math.max(corner[0].y, corner[1].y)); } public void set(RectF rect) { set(new Vector2D(rect.centerX(),rect.centerY()),rect.width(),rect.height(),0.0f); } // Returns true if other overlaps one dimension of this. private boolean overlaps1Way(OBB2D other) { for (int a = 0; a < axis.length; ++a) { double t = other.corner[0].dot(axis[a]); // Find the extent of box 2 on axis a double tMin = t; double tMax = t; for (int c = 1; c < corner.length; ++c) { t = other.corner[c].dot(axis[a]); if (t < tMin) { tMin = t; } else if (t > tMax) { tMax = t; } } // We have to subtract off the origin // See if [tMin, tMax] intersects [0, 1] if ((tMin > 1 + origin[a]) || (tMax < origin[a])) { // There was no intersection along this dimension; // the boxes cannot possibly overlap. return false; } } // There was no dimension along which there is no intersection. // Therefore the boxes overlap. return true; } //Updates the axes after the corners move. Assumes the //corners actually form a rectangle. private void computeAxes() { axis[0] = corner[1].subtract(corner[0]); axis[1] = corner[3].subtract(corner[0]); // Make the length of each axis 1/edge length so we know any // dot product must be less than 1 to fall within the edge. for (int a = 0; a < axis.length; ++a) { axis[a] = axis[a].divide((axis[a].length() * axis[a].length())); origin[a] = corner[0].dot(axis[a]); } } public void moveTo(Vector2D center) { Vector2D centroid = (corner[0].add(corner[1]).add(corner[2]).add(corner[3])).divide(4.0f); Vector2D translation = center.subtract(centroid); for (int c = 0; c < 4; ++c) { corner[c] = corner[c].add(translation); } computeAxes(); computeDimensions(center,angle); } // Returns true if the intersection of the boxes is non-empty. public boolean overlaps(OBB2D other) { if(right() < other.left()) { return false; } if(bottom() < other.top()) { return false; } if(left() > other.right()) { return false; } if(top() > other.bottom()) { return false; } if(other.getAngle() == 0.0f && getAngle() == 0.0f) { return true; } return overlaps1Way(other) && other.overlaps1Way(this); } public Vector2D getCenter() { return center; } public float getWidth() { return extents.x * 2; } public float getHeight() { return extents.y * 2; } public void setAngle(float angle) { set(center,getWidth(),getHeight(),angle); } public float getAngle() { return angle; } public void setSize(float w,float h) { set(center,w,h,angle); } public float left() { return boundingRect.left; } public float right() { return boundingRect.right; } public float bottom() { return boundingRect.bottom; } public float top() { return boundingRect.top; } public RectF getBoundingRect() { return boundingRect; } public boolean overlaps(float left, float top, float right, float bottom) { if(right() < left) { return false; } if(bottom() < top) { return false; } if(left() > right) { return false; } if(top() > bottom) { return false; } return true; } };

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  • South migration error: NoMigrations exception for django.contrib.auth

    - by danpalmer
    I have been using South on my project for a while, but I recently did a huge amount of development and changed development machine and I think something messed up in the process. The project works fine, but I can't apply migrations. Whenever I try to apply a migration I get the following traceback: danpalmer:pest Dan$ python manage.py migrate frontend Traceback (most recent call last): File "manage.py", line 11, in <module> execute_manager(settings) File "/Library/Python/2.6/site-packages/django/core/management/__init__.py", line 362, in execute_manager utility.execute() File "/Library/Python/2.6/site-packages/django/core/management/__init__.py", line 303, in execute self.fetch_command(subcommand).run_from_argv(self.argv) File "/Library/Python/2.6/site-packages/django/core/management/base.py", line 195, in run_from_argv self.execute(*args, **options.__dict__) File "/Library/Python/2.6/site-packages/django/core/management/base.py", line 222, in execute output = self.handle(*args, **options) File "/Library/Python/2.6/site-packages/South-0.7-py2.6.egg/south/management/commands/migrate.py", line 102, in handle delete_ghosts = delete_ghosts, File "/Library/Python/2.6/site-packages/South-0.7-py2.6.egg/south/migration/__init__.py", line 182, in migrate_app applied = check_migration_histories(applied, delete_ghosts) File "/Library/Python/2.6/site-packages/South-0.7-py2.6.egg/south/migration/__init__.py", line 85, in check_migration_histories m = h.get_migration() File "/Library/Python/2.6/site-packages/South-0.7-py2.6.egg/south/models.py", line 34, in get_migration return self.get_migrations().migration(self.migration) File "/Library/Python/2.6/site-packages/South-0.7-py2.6.egg/south/models.py", line 31, in get_migrations return Migrations(self.app_name) File "/Library/Python/2.6/site-packages/South-0.7-py2.6.egg/south/migration/base.py", line 60, in __call__ self.instances[app_label] = super(MigrationsMetaclass, self).__call__(app_label_to_app_module(app_label), **kwds) File "/Library/Python/2.6/site-packages/South-0.7-py2.6.egg/south/migration/base.py", line 88, in __init__ self.set_application(application, force_creation, verbose_creation) File "/Library/Python/2.6/site-packages/South-0.7-py2.6.egg/south/migration/base.py", line 159, in set_application raise exceptions.NoMigrations(application) south.exceptions.NoMigrations: Application '<module 'django.contrib.auth' from '/Library/Python/2.6/site-packages/django/contrib/auth/__init__.pyc'>' has no migrations. I am not that experienced with South and I haven't met this error before. The only helpful mention I can find online about this error is for pre-0.7 I think and I am on South 0.7. I ran 'easy_install -U South' just to make sure. Thanks for any help that you can provide. I really appreciate it.

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  • Java: initialization problem with private-final-int-value and empty constructor

    - by HH
    $ javac InitInt.java InitInt.java:7: variable right might not have been initialized InitInt(){} ^ 1 error $ cat InitInt.java import java.util.*; import java.io.*; public class InitInt { private final int right; InitInt(){} public static void main(String[] args) { // I don't want to assign any value. // just initialize it, how? InitInt test = new InitInt(); System.out.println(test.getRight()); // later assiging a value } public int getRight(){return right;} } Initialization problem with Constructor InitInt{ // Still the error, "may not be initialized" // How to initialise it? if(snippetBuilder.length()>(charwisePos+25)){ right=charwisePos+25; }else{ right=snippetBuilder.length()-1; } }

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  • Off center projection

    - by N0xus
    I'm trying to implement the code that was freely given by a very kind developer at the following link: http://forum.unity3d.com/threads/142383-Code-sample-Off-Center-Projection-Code-for-VR-CAVE-or-just-for-fun Right now, all I'm trying to do is bring it in on one camera, but I have a few issues. My class, looks as follows: using UnityEngine; using System.Collections; public class PerspectiveOffCenter : MonoBehaviour { // Use this for initialization void Start () { } // Update is called once per frame void Update () { } public static Matrix4x4 GeneralizedPerspectiveProjection(Vector3 pa, Vector3 pb, Vector3 pc, Vector3 pe, float near, float far) { Vector3 va, vb, vc; Vector3 vr, vu, vn; float left, right, bottom, top, eyedistance; Matrix4x4 transformMatrix; Matrix4x4 projectionM; Matrix4x4 eyeTranslateM; Matrix4x4 finalProjection; ///Calculate the orthonormal for the screen (the screen coordinate system vr = pb - pa; vr.Normalize(); vu = pc - pa; vu.Normalize(); vn = Vector3.Cross(vr, vu); vn.Normalize(); //Calculate the vector from eye (pe) to screen corners (pa, pb, pc) va = pa-pe; vb = pb-pe; vc = pc-pe; //Get the distance;; from the eye to the screen plane eyedistance = -(Vector3.Dot(va, vn)); //Get the varaibles for the off center projection left = (Vector3.Dot(vr, va)*near)/eyedistance; right = (Vector3.Dot(vr, vb)*near)/eyedistance; bottom = (Vector3.Dot(vu, va)*near)/eyedistance; top = (Vector3.Dot(vu, vc)*near)/eyedistance; //Get this projection projectionM = PerspectiveOffCenter(left, right, bottom, top, near, far); //Fill in the transform matrix transformMatrix = new Matrix4x4(); transformMatrix[0, 0] = vr.x; transformMatrix[0, 1] = vr.y; transformMatrix[0, 2] = vr.z; transformMatrix[0, 3] = 0; transformMatrix[1, 0] = vu.x; transformMatrix[1, 1] = vu.y; transformMatrix[1, 2] = vu.z; transformMatrix[1, 3] = 0; transformMatrix[2, 0] = vn.x; transformMatrix[2, 1] = vn.y; transformMatrix[2, 2] = vn.z; transformMatrix[2, 3] = 0; transformMatrix[3, 0] = 0; transformMatrix[3, 1] = 0; transformMatrix[3, 2] = 0; transformMatrix[3, 3] = 1; //Now for the eye transform eyeTranslateM = new Matrix4x4(); eyeTranslateM[0, 0] = 1; eyeTranslateM[0, 1] = 0; eyeTranslateM[0, 2] = 0; eyeTranslateM[0, 3] = -pe.x; eyeTranslateM[1, 0] = 0; eyeTranslateM[1, 1] = 1; eyeTranslateM[1, 2] = 0; eyeTranslateM[1, 3] = -pe.y; eyeTranslateM[2, 0] = 0; eyeTranslateM[2, 1] = 0; eyeTranslateM[2, 2] = 1; eyeTranslateM[2, 3] = -pe.z; eyeTranslateM[3, 0] = 0; eyeTranslateM[3, 1] = 0; eyeTranslateM[3, 2] = 0; eyeTranslateM[3, 3] = 1f; //Multiply all together finalProjection = new Matrix4x4(); finalProjection = Matrix4x4.identity * projectionM*transformMatrix*eyeTranslateM; //finally return return finalProjection; } // Update is called once per frame public void FixedUpdate () { Camera cam = camera; //calculate projection Matrix4x4 genProjection = GeneralizedPerspectiveProjection( new Vector3(0,1,0), new Vector3(1,1,0), new Vector3(0,0,0), new Vector3(0,0,0), cam.nearClipPlane, cam.farClipPlane); //(BottomLeftCorner, BottomRightCorner, TopLeftCorner, trackerPosition, cam.nearClipPlane, cam.farClipPlane); cam.projectionMatrix = genProjection; } } My error lies in projectionM = PerspectiveOffCenter(left, right, bottom, top, near, far); The debugger states: Expression denotes a `type', where a 'variable', 'value' or 'method group' was expected. Thus, I changed the line to read: projectionM = new PerspectiveOffCenter(left, right, bottom, top, near, far); But then the error is changed to: The type 'PerspectiveOffCenter' does not contain a constructor that takes '6' arguments. For reasons that are obvious. So, finally, I changed the line to read: projectionM = new GeneralizedPerspectiveProjection(left, right, bottom, top, near, far); And the error I get is: is a 'method' but a 'type' was expected. With this last error, I'm not sure what it is I should do / missing. Can anyone see what it is that I'm missing to fix this error?

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  • jQuery to get innerHTML not working on a HTMLFontElement object...

    - by Matt
    I have jQuery to select all font elements that are children of the element with id="right" within the html stored in the var html... when I do an alert to see how many elements it gets: alert($("#right > font", html).length); it gives me an alert of: 5 but when I try any of the following, I don't get any alerts... alert($("#right > font", html)[0].html()); alert($("#right > font", html)[0].text()); alert($("#right > font", html)[0].val()); Any Ideas? Thanks, Matt

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  • How to occupy all the space in a div when working with min-height header / footer

    - by javacoder
    I believe this is a beginner's CSS question. I am utilizing the method described in http://www.xs4all.nl/~peterned/examples/csslayout1.html to fix a header to the top and a footer to the bottom. What I'd like to achieve now is two columns inside the content div. A left one of 200px and a right one that takes up the rest of the width. Unfortunately, I can't get the left and right divs to display correctly: they just don't grow vertically, and if I make the right div "width: 100%" it positions itself underneath the left one. What is the trick to make the left and right div take up all the space within the content div? The layout1.css is the original one. I just added two entries: #left and #right layout1.css: /** * 100% height layout with header and footer * ---------------------------------------------- * Feel free to copy/use/change/improve */ html,body { margin: 0; padding: 0; height: 100%; /* needed for container min-height */ background: gray; font-family: arial, sans-serif; font-size: small; color: #666; } h1 { font: 1.5em georgia, serif; margin: 0.5em 0; } h2 { font: 1.25em georgia, serif; margin: 0 0 0.5em; } h1,h2,a { color: orange; } p { line-height: 1.5; margin: 0 0 1em; } div#container { position: relative; /* needed for footer positioning*/ margin: 0 auto; /* center, not in IE5 */ width: 750px; background: #f0f0f0; height: auto !important; /* real browsers */ height: 100%; /* IE6: treaded as min-height*/ min-height: 100%; /* real browsers */ } div#header { padding: 1em; background: #ddd url("../csslayout.gif") 98% 10px no-repeat; border-bottom: 6px double gray; } div#header p { font-style: italic; font-size: 1.1em; margin: 0; } div#content { padding: 1em 1em 5em; /* bottom padding for footer */ } div#content p { text-align: justify; padding: 0 1em; } div#footer { position: absolute; width: 100%; bottom: 0; /* stick to bottom */ background: #ddd; border-top: 6px double gray; } div#footer p { padding: 1em; margin: 0; } // added the following: div#left { border: 1px solid red; width: 200px; float: left; min-height: 100%; height: 100%; padding-left: 10px; margin-right: 10px; } div#right { border: 1px solid blue; float: left; min-height: 100%; height: 100%; padding-left: 10px; margin-right: 10px; } layout.html: <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>CSS Layout - 100% height</title> <link rel="stylesheet" type="text/css" href="layout1.css" /> </head> <body> <div id="container"> <div id="header"> <h1>header</h1> </div> <div id="content"> <div id="left"> left column </div> <div id="right"> right column </div> </div> <div id="footer"> <p> footer </p> </div> </div> </body>

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  • Java: design problem with final-value and empty constructor

    - by HH
    $ javac InitInt.java InitInt.java:7: variable right might not have been initialized InitInt(){} ^ 1 error $ cat InitInt.java import java.util.*; import java.io.*; public class InitInt { private final int right; // Design Problem? // I feel the initialization problem is just due to bad style. InitInt(){} InitInt{ // Still the error, "may not be initialized" // How to initialise it? if(snippetBuilder.length()>(charwisePos+25)){ right=charwisePos+25; }else{ right=snippetBuilder.length()-1; } } public static void main(String[] args) { InitInt test = new InitInt(); System.out.println(test.getRight()); } public int getRight(){return right;} } Partial Solutions and Suggestions use "this" to access methods in the class, instead of creating empty constructor change final to non-final with final field value: initialize all final values in every constructor remove the empty constructor, keep your code simple and clean

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  • Top 10 Browser Productivity Tips

    - by Renso
    Originally posted on: http://geekswithblogs.net/renso/archive/2013/10/14/top-10-browser-productivity-tips.aspxYou don’t have to be a geek to be a productive browser user. The tips below have been selected by actions users take most of the time to navigate a web-site but use long-standing keyboard or mouse actions to get them done, when there are keyboard short-cuts you can use instead. Since you hands are already on the keyboard it is almost always faster to sue a keyboard shortcut to get something done that you usually used the mouse for. For example right-clicking on something to copy it and then doing to same for pasting something is very time consuming, keyboard shortcuts have been created that simplify the task. All it takes are a few memory brain cells to remember them. Here are the tips, in no particular order:   Tip 1 Hold down the spacebar on your keyboard to page to the end of your web page rather than using your mouse. This is really a slow way of doing it. If you want to page one page at a time, hit the spacebar once, and again to page again. But if you want to page all the way to the end of the web page simply hit Ctrl+End (that is hold down the Ctrl key and hit the End button on your keyboard). To get to the top of your web page, simply hit Ctrl + Home to go all the way to the top of your web page. Tip 2 Where are my downloads? Some folks run downloads again-and-again because they do not know where the last one went and they do not see the popup, or browser note on their web page in the footer, etc. Simply hit Ctrl+J. Works in most browsers. Tip 3 Selecting a US state from a drop down box. Don’t use the mouse, takes just way too long to scroll. When you tab to the drop down box or click on it with your mouse, simply hit the first character of the state and it will be selected. For Texas for example hit the letter “T” twice on your keyboard to get to it. The same concept can be applied to any drop down box that is alphabetical or numerically sorted. Tip 4 Fixing spelling errors. All modern-day browsers support this now. You see the red wavy lines underscoring a word, yes it is a spelling error. How do you fix it? Don’t overtype it or try and fix it manually, fist right-click on it and a list of suggestions comes up. If it does not show up, like my name “Renso” and you know how to spell your name as in this example, look further down the list of options (the little window popup that appears when you right click) and you should see an option to “Add to Dictionary”. Be warned, when you add it, it only adds it to the browser you’re using’s dictionary. If you use Google Chrome, Firefox and IE, each one will have their own list. Tip 5 So you have trouble seeing the text on the screen. Or you are looking at a photo, for example in Facebook. You want to zoom in to read better or zoom into a photo a bit more. Hit Ctrl++ (hold down Ctrl key and hit the plus key – actually it’s the equal key but it is easier to remember that it is plus for bigger). Hit the minus to zoom out. Now you can’t remember what the original size was since you were so excited to hit it 20 times, or was that 21… Simply hit Ctrl+0 (that is zero) and it will reset it to the default. Tip 6 So you closed a couple of tabs in your browser. Suddenly you remember something you wanted to double-check something on one of the tabs, you cannot remember the URL ad the tab is gone forever, or is it? Simply hit Ctrl+Shift+t and it will bring back your tabs one by one each time you click the T. This has also been a great way for me to quickly close some tabs because I don’t want my boss to see I’m shopping and then hitting Ctrl+Shift+t to quickly get it back and complete my check-put and purchase. Or, for parents, when you walk into your daughter’s room and you see she quickly clicks and closes a window/tab in here browser. Not to worry my little darling, daddy will Ctrl+Shift+t and see what boys on Facebook you were talking too… Tip 7 The web browser is frozen on your PC/Laptop/Whatever, in this example it may be your Internet Explorer browser. I don’t mention Firefox or Chrome here because it probably never happens in their world. You cannot close it, it won’t respond to anything you have done s far except for the next step you are about to take, which is throw your two-day old coffee on your keyboard. This happens especially on sites that want to force you to complete a purchase order. Hit Ctrl+Alt+Del on your keyboard on any version of windows, select TASK MANAGER. In the  First Tab, which is the Process Tab, look for the item in question. In this example you should see Internet Explorer. Right-click it and select “End Task”. It will force the thread out of memory and terminate that process. You can of course do this with any program running under your account. Tip 8 This is a personal favorite of mine. To select words in the paragraph without using the mouse. You don’t want to select one character at a time like when you use the Ctrl+arrows as it can be very slow if you want to select a lot of text. You also want to select whole words. Simply use the Ctrl+Shift_arrow (right or left depending which direction you want to go. Tip 9 I was a bit reluctant to add this one, but being in the professional services industry still come across many-a-folk that simply can’t copy-and-paste them-all text or images that reside on them screens, y’all. Ctrl+c to copy and Ctrl+v to paste it. Works a lot faster than using the mouse. You may be asking: “Well why in the devil did they not use Ctrl+p for paste…. because that is for printing. This is of course not limited to the browser world, it applies to almost any piece of software running on PC or Mac. Go try it on an image on your browser, right-click it and select copy. Open a word document and Ctrl+v to paste the image in there. Please consider copyright laws. Tip 10 Getting rid of annoying ads. Now this only works when you load a web page, meaning when you get back to the same page later you will have to do this again and you will need to learn a tool to do it, WELL WORTH IT. For example, I use GrooveShark to listen to music but I don’t like the ads they show. Install a tool like Firebug for Firefox or use the Ctrl+Shift+I on Chrome to bring up the developer toolbar. Shows at the bottom of the page. With Firefox, once you have installed Firebug as an add-on, a yellow bug should appear on the top right-hand-side of your browser, click on it to display the developer toolbar. You will need to learn how to use it, but once you know how to select an item/section on the window (usually just right-click the add you don’t want to see and select “Inspect Element”, the developer toolbar will appear (if not already there)) and then simply hit delete and it will remove the add from the screen. If you don’t know HTML you may need to play with it a bit, but once you understand how it works can open up a whole new world for you on how web pages actually work. If you can think of any others that have saved you a ton of time please let me know so I can add them to a top 99 list.

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  • Java: how to initialize private final int value with if-else in constructor?

    - by HH
    $ javac InitInt.java InitInt.java:7: variable right might not have been initialized InitInt(){} ^ 1 error $ cat InitInt.java import java.util.*; import java.io.*; public class InitInt { private final int right; InitInt(){} public static void main(String[] args) { // I don't want to assign any value. // just initialize it, how? InitInt test = new InitInt(); System.out.println(test.getRight()); // later assiging a value } public int getRight(){return right;} } Initialization problem with Constructor, due to if-else -loop InitInt{ // Still the error, "may not be initialized" // How to initialise it, without removing if-else? if(snippetBuilder.length()>(charwisePos+25)){ right=charwisePos+25; }else{ right=snippetBuilder.length()-1; } }

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  • Displaying UserControls based on the type a TreeView selection is bound to

    - by Ray Wenderlich
    I am making an app in WPF in a style similar to Windows Explorer, with a TreeView on the left and a pane on the right. I want the contents of the right pane to change depending on the type of the selected element in the TreeView. For example, say the top level in the Tree View contains objects of class "A", and if you expand the "A" object you'll see a list of "B" objects as children of the "A" object. If the "A" object is selected, I want the right pane to show a user control for "A", and if "B" is selected I want the right pane to show a user control for "B". I've currently got this working by: setting up the TreeView with one HierarchialDataTemplate per type adding all the UserControls to the right pane, but collapsed implementing SelectedItemChanged on the TreeView, and setting the appropriate usercontrol to visible and the others to collapsed. However, I'm sure there's a better/more elegant way to switch out the views based on the type the selection is bound to, perhaps by making more use of data binding... any ideas?

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  • Finding the heaviest length-constrained path in a weighted Binary Tree

    - by Hristo
    UPDATE I worked out an algorithm that I think runs in O(n*k) running time. Below is the pseudo-code: routine heaviestKPath( T, k ) // create 2D matrix with n rows and k columns with each element = -8 // we make it size k+1 because the 0th column must be all 0s for a later // function to work properly and simplicity in our algorithm matrix = new array[ T.getVertexCount() ][ k + 1 ] (-8); // set all elements in the first column of this matrix = 0 matrix[ n ][ 0 ] = 0; // fill our matrix by traversing the tree traverseToFillMatrix( T.root, k ); // consider a path that would arc over a node globalMaxWeight = -8; findArcs( T.root, k ); return globalMaxWeight end routine // node = the current node; k = the path length; node.lc = node’s left child; // node.rc = node’s right child; node.idx = node’s index (row) in the matrix; // node.lc.wt/node.rc.wt = weight of the edge to left/right child; routine traverseToFillMatrix( node, k ) if (node == null) return; traverseToFillMatrix(node.lc, k ); // recurse left traverseToFillMatrix(node.rc, k ); // recurse right // in the case that a left/right child doesn’t exist, or both, // let’s assume the code is smart enough to handle these cases matrix[ node.idx ][ 1 ] = max( node.lc.wt, node.rc.wt ); for i = 2 to k { // max returns the heavier of the 2 paths matrix[node.idx][i] = max( matrix[node.lc.idx][i-1] + node.lc.wt, matrix[node.rc.idx][i-1] + node.rc.wt); } end routine // node = the current node, k = the path length routine findArcs( node, k ) if (node == null) return; nodeMax = matrix[node.idx][k]; longPath = path[node.idx][k]; i = 1; j = k-1; while ( i+j == k AND i < k ) { left = node.lc.wt + matrix[node.lc.idx][i-1]; right = node.rc.wt + matrix[node.rc.idx][j-1]; if ( left + right > nodeMax ) { nodeMax = left + right; } i++; j--; } // if this node’s max weight is larger than the global max weight, update if ( globalMaxWeight < nodeMax ) { globalMaxWeight = nodeMax; } findArcs( node.lc, k ); // recurse left findArcs( node.rc, k ); // recurse right end routine Let me know what you think. Feedback is welcome. I think have come up with two naive algorithms that find the heaviest length-constrained path in a weighted Binary Tree. Firstly, the description of the algorithm is as follows: given an n-vertex Binary Tree with weighted edges and some value k, find the heaviest path of length k. For both algorithms, I'll need a reference to all vertices so I'll just do a simple traversal of the Tree to have a reference to all vertices, with each vertex having a reference to its left, right, and parent nodes in the tree. Algorithm 1 For this algorithm, I'm basically planning on running DFS from each node in the Tree, with consideration to the fixed path length. In addition, since the path I'm looking for has the potential of going from left subtree to root to right subtree, I will have to consider 3 choices at each node. But this will result in a O(n*3^k) algorithm and I don't like that. Algorithm 2 I'm essentially thinking about using a modified version of Dijkstra's Algorithm in order to consider a fixed path length. Since I'm looking for heaviest and Dijkstra's Algorithm finds the lightest, I'm planning on negating all edge weights before starting the traversal. Actually... this doesn't make sense since I'd have to run Dijkstra's on each node and that doesn't seem very efficient much better than the above algorithm. So I guess my main questions are several. Firstly, do the algorithms I've described above solve the problem at hand? I'm not totally certain the Dijkstra's version will work as Dijkstra's is meant for positive edge values. Now, I am sure there exist more clever/efficient algorithms for this... what is a better algorithm? I've read about "Using spine decompositions to efficiently solve the length-constrained heaviest path problem for trees" but that is really complicated and I don't understand it at all. Are there other algorithms that tackle this problem, maybe not as efficiently as spine decomposition but easier to understand? Thanks.

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  • liquid CSS issues - rtl, floating and scrollers

    - by Rani
    hi I want to build a site that will have these restrictions: RTL direction vertical scroll on the right side whole page is floating to the right page has 2 columns the right (main) column has min width the right (main) column has table inside it that can expend in its data and get wider making all other data in the column expend to the same width as well the sidebar should be on the left side but still floating to the right of the main div it should fit low resolution so the page will be able to add horizontal scroll if needed should work in all major browsers don't use table for constructing the page Can someone help or direct me? Thanks Rani

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  • How do implement a breadth first traversal?

    - by not looking for answer
    //This is what I have. I thought pre-order was the same and mixed it up with depth first! import java.util.LinkedList; import java.util.Queue; public class Exercise25_1 { public static void main(String[] args) { BinaryTree tree = new BinaryTree(new Integer[] {10, 5, 15, 12, 4, 8 }); System.out.print("\nInorder: "); tree.inorder(); System.out.print("\nPreorder: "); tree.preorder(); System.out.print("\nPostorder: "); tree.postorder(); //call the breadth method to test it System.out.print("\nBreadthFirst:"); tree.breadth(); } } class BinaryTree { private TreeNode root; /** Create a default binary tree */ public BinaryTree() { } /** Create a binary tree from an array of objects */ public BinaryTree(Object[] objects) { for (int i = 0; i < objects.length; i++) { insert(objects[i]); } } /** Search element o in this binary tree */ public boolean search(Object o) { return search(o, root); } public boolean search(Object o, TreeNode root) { if (root == null) { return false; } if (root.element.equals(o)) { return true; } else { return search(o, root.left) || search(o, root.right); } } /** Return the number of nodes in this binary tree */ public int size() { return size(root); } public int size(TreeNode root) { if (root == null) { return 0; } else { return 1 + size(root.left) + size(root.right); } } /** Return the depth of this binary tree. Depth is the * number of the nodes in the longest path of the tree */ public int depth() { return depth(root); } public int depth(TreeNode root) { if (root == null) { return 0; } else { return 1 + Math.max(depth(root.left), depth(root.right)); } } /** Insert element o into the binary tree * Return true if the element is inserted successfully */ public boolean insert(Object o) { if (root == null) { root = new TreeNode(o); // Create a new root } else { // Locate the parent node TreeNode parent = null; TreeNode current = root; while (current != null) { if (((Comparable)o).compareTo(current.element) < 0) { parent = current; current = current.left; } else if (((Comparable)o).compareTo(current.element) > 0) { parent = current; current = current.right; } else { return false; // Duplicate node not inserted } } // Create the new node and attach it to the parent node if (((Comparable)o).compareTo(parent.element) < 0) { parent.left = new TreeNode(o); } else { parent.right = new TreeNode(o); } } return true; // Element inserted } public void breadth() { breadth(root); } // Implement this method to produce a breadth first // search traversal public void breadth(TreeNode root){ if (root == null) return; System.out.print(root.element + " "); breadth(root.left); breadth(root.right); } /** Inorder traversal */ public void inorder() { inorder(root); } /** Inorder traversal from a subtree */ private void inorder(TreeNode root) { if (root == null) { return; } inorder(root.left); System.out.print(root.element + " "); inorder(root.right); } /** Postorder traversal */ public void postorder() { postorder(root); } /** Postorder traversal from a subtree */ private void postorder(TreeNode root) { if (root == null) { return; } postorder(root.left); postorder(root.right); System.out.print(root.element + " "); } /** Preorder traversal */ public void preorder() { preorder(root); } /** Preorder traversal from a subtree */ private void preorder(TreeNode root) { if (root == null) { return; } System.out.print(root.element + " "); preorder(root.left); preorder(root.right); } /** Inner class tree node */ private class TreeNode { Object element; TreeNode left; TreeNode right; public TreeNode(Object o) { element = o; } } }

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  • make tree in scheme

    - by ???
    (define (entry tree) (car tree)) (define (left-branch tree) (cadr tree)) (define (right-branch tree) (caddr tree)) (define (make-tree entry left right) (list entry left right)) (define (mktree order items_list) (cond ((= (length items_list) 1) (make-tree (car items_list) '() '())) (else (insert2 order (car items_list) (mktree order (cdr items_list)))))) (define (insert2 order x t) (cond ((null? t) (make-tree x '() '())) ((order x (entry t)) (make-tree (entry t) (insert2 order x (left-branch t)) (right-branch t))) ((order (entry t) x ) (make-tree (entry t) (left-branch t) (insert2 order x (right-branch t)))) (else t))) The result is: (mktree (lambda (x y) (< x y)) (list 7 3 5 1 9 11)) (11 (9 (1 () (5 (3 () ()) (7 () ()))) ()) ()) But I'm trying to get: (7 (3 (1 () ()) (5 () ())) (9 () (11 () ()))) Where is the problem?

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  • Placing an background image with padding in h2 tag

    - by Cedar Jensen
    I want to create a headline (h2) with an image at the right-most area of the bounding box. I have the layout almost right except I can't push the image a little bit to the right of the element's bounding box -- how would I tweak my css so it is displayed correctly? I'm trying to do something like this: [{someHeadLineText}{dynamic space }{image}{5px space}] where the [] indicate the total available width of my content. Html: <div class="primaryHeader"> <h2>News</h2> </div> Css: .primaryHeader h2 { background-color: green; /* the header looks like a box */ color: black; background: transparent url(../images/edit.png) no-repeat right center; border: 1px solid red; } I am placing the image to the right of my h2 element and centered vertically -- but how do I adjust the placement of the background image?

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  • In 3D camera math, calculate what Z depth is pixel unity for a given FOV

    - by badweasel
    I am working in iOS and OpenGL ES 2.0. Through trial and error I've figured out a frustum to where at a specific z depth pixels drawn are 1 to 1 with my source textures. So 1 pixel in my texture is 1 pixel on the screen. For 2d games this is good. Of course it means that I also factor in things like the size of the quad and the size of the texture. For example if my sprite is a quad 32x32 pixels. The quad size is 3.2 units wide and tall. And the texcoords are 32 / the size of the texture wide and tall. Then the frustum is: matrixFrustum(-(float)backingWidth/frustumScale,(float)backingWidth/frustumScale, -(float)backingHeight/frustumScale, (float)backingHeight/frustumScale, 40, 1000, mProjection); Where frustumScale is 800 for a retina screen. Then at a distance of 800 from camera the sprite is pixel for pixel the same as photoshop. For 3d games sometimes I still want to be able to do this. But depending on the scene I sometimes need the FOV to be different things. I'm looking for a way to figure out what Z depth will achieve this same pixel unity for a given FOV. For this my mProjection is set using: matrixPerspective(cameraFOV, near, far, (float)backingWidth / (float)backingHeight, mProjection); With testing I found that at an FOV of 45.0 a Z of 38.5 is very close to pixel unity. And at an FOV of 30.0 a Z of 59.5 is about right. But how can I calculate a value that is spot on? Here's my matrixPerspecitve code: void matrixPerspective(float angle, float near, float far, float aspect, mat4 m) { //float size = near * tanf(angle / 360.0 * M_PI); float size = near * tanf(degreesToRadians(angle) / 2.0); float left = -size, right = size, bottom = -size / aspect, top = size / aspect; // Unused values in perspective formula. m[1] = m[2] = m[3] = m[4] = 0; m[6] = m[7] = m[12] = m[13] = m[15] = 0; // Perspective formula. m[0] = 2 * near / (right - left); m[5] = 2 * near / (top - bottom); m[8] = (right + left) / (right - left); m[9] = (top + bottom) / (top - bottom); m[10] = -(far + near) / (far - near); m[11] = -1; m[14] = -(2 * far * near) / (far - near); } And my mView is set using: lookAtMatrix(cameraPos, camLookAt, camUpVector, mView); * UPDATE * I'm going to leave this here in case anyone has a different solution, can explain how they do it, or why this works. This is what I figured out. In my system I use a 10th scale unit to pixels on non-retina displays and a 20th scale on retina displays. The iPhone is 640 pixels wide on retina and 320 pixels wide on non-retina (obsolete). So if I want something to be the full screen width I divide by 20 to get the OpenGL unit width. Then divide that by 2 to get the left and right unit position. Something 32 units wide centered on the screen goes from -16 to +16. Believe it or not I have an excel spreadsheet do all this math for me and output all the vertex data for my sprite sheet. It's an arbitrary thing I made up to do .1 units = 1 non-retina pixel or 2 retina pixels. I could have made it .01 units = 2 pixels and someday I might switch to that. But for now it's the other. So the width of the screen in units is 32.0, and that means the left most pixel is at -16.0 and the right most is at 16.0. After messing a bit I figured out that if I take the [0] value of an identity modelViewProjection matrix and multiply it by 16 I get the depth required to get 1:1 pixels. I don't know why. I don't know if the 16 is related to the screen size or just a lucky guess. But I did a test where I placed a sprite at that calculated depth and varied the FOV through all the valid values and the object stays steady on screen with 1:1 pixels. So now I'm just calculating the unityDepth that way. If someone gives me a better answer I'll checkmark it.

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  • Trouble using South with Django and Heroku

    - by Dan
    I had an existing Django project that I've just added South to. I ran syncdb locally. I ran manage.py schemamigration app_name locally I ran manage.py migrate app_name --fake locally I commit and pushed to heroku master I ran syncdb on heroku I ran manage.py schemamigration app_name on heroku I ran manage.py migrate app_name on heroku I then receive this: $ heroku run python notecard/manage.py migrate notecards Running python notecard/manage.py migrate notecards attached to terminal... up, run.1 Running migrations for notecards: - Migrating forwards to 0005_initial. > notecards:0003_initial Traceback (most recent call last): File "notecard/manage.py", line 14, in <module> execute_manager(settings) File "/app/lib/python2.7/site-packages/django/core/management/__init__.py", line 438, in execute_manager utility.execute() File "/app/lib/python2.7/site-packages/django/core/management/__init__.py", line 379, in execute self.fetch_command(subcommand).run_from_argv(self.argv) File "/app/lib/python2.7/site-packages/django/core/management/base.py", line 191, in run_from_argv self.execute(*args, **options.__dict__) File "/app/lib/python2.7/site-packages/django/core/management/base.py", line 220, in execute output = self.handle(*args, **options) File "/app/lib/python2.7/site-packages/south/management/commands/migrate.py", line 105, in handle ignore_ghosts = ignore_ghosts, File "/app/lib/python2.7/site-packages/south/migration/__init__.py", line 191, in migrate_app success = migrator.migrate_many(target, workplan, database) File "/app/lib/python2.7/site-packages/south/migration/migrators.py", line 221, in migrate_many result = migrator.__class__.migrate_many(migrator, target, migrations, database) File "/app/lib/python2.7/site-packages/south/migration/migrators.py", line 292, in migrate_many result = self.migrate(migration, database) File "/app/lib/python2.7/site-packages/south/migration/migrators.py", line 125, in migrate result = self.run(migration) File "/app/lib/python2.7/site-packages/south/migration/migrators.py", line 99, in run return self.run_migration(migration) File "/app/lib/python2.7/site-packages/south/migration/migrators.py", line 81, in run_migration migration_function() File "/app/lib/python2.7/site-packages/south/migration/migrators.py", line 57, in <lambda> return (lambda: direction(orm)) File "/app/notecard/notecards/migrations/0003_initial.py", line 15, in forwards ('user', self.gf('django.db.models.fields.related.ForeignKey')(to=orm['auth.User'])), File "/app/lib/python2.7/site-packages/south/db/generic.py", line 226, in create_table ', '.join([col for col in columns if col]), File "/app/lib/python2.7/site-packages/south/db/generic.py", line 150, in execute cursor.execute(sql, params) File "/app/lib/python2.7/site-packages/django/db/backends/util.py", line 34, in execute return self.cursor.execute(sql, params) File "/app/lib/python2.7/site-packages/django/db/backends/postgresql_psycopg2/base.py", line 44, in execute return self.cursor.execute(query, args) django.db.utils.DatabaseError: relation "notecards_semester" already exists I have 3 models. Section, Semester, and Notecards. I've added one field to the Notecards model and I cannot get it added on Heroku. Thank you.

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  • Java: design problem with private-final-int-value and empty constructor

    - by HH
    $ javac InitInt.java InitInt.java:7: variable right might not have been initialized InitInt(){} ^ 1 error $ cat InitInt.java import java.util.*; import java.io.*; public class InitInt { private final int right; //DUE to new Klowledge: Design Problem //I think having an empty constructor like this // is an design problem, shall I remove it? What do you think? // When to use an empty constructor? InitInt(){} public static void main(String[] args) { InitInt test = new InitInt(); System.out.println(test.getRight()); } public int getRight(){return right;} } Initialization problem with Constructor InitInt{ // Still the error, "may not be initialized" // How to initialise it? if(snippetBuilder.length()>(charwisePos+25)){ right=charwisePos+25; }else{ right=snippetBuilder.length()-1; } }

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  • How do I make the footer stretch vertically downward align to footer.

    - by text
    I am new to web design using tableless and I'm having problem positioning some elements on my page.. Here's the sample html: http://christianruado.comuf.com/sample.html As you can see from the screen shots I want my right div to be vertically stretched down to the same level of my footer and position my bottom element to the lowest part of the right container. #right { float:right; width: 19%; background:#FF3300; margin-left:2px; height: 100%; } #right .bottom { bottom:0; display:block; background-color:#FFCCFF; height:30px; }

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  • Drupal menu permissions question

    - by Luke
    I'm creating an admin module for my client that gives then access to some administration functionality concerning their content. I'm starting off my adding some permissions in my module by implementing hook_perm: function mymodule_perm() { return array( 'manage projects', ); } I can then create my menu by adding to the admin section that already exists: function mymodule_menu() { $items['admin/projects'] = array( 'title' => 'Projects', 'description' => 'Manage your projects.', 'page callback' => 'manage_projects_overview', 'access callback' => 'user_access', 'access arguments' => array('manage projects'), 'type' => MENU_NORMAL_ITEM, 'weight' => -100, ); $items['admin/projects/add'] = array( 'title' => 'Add project', 'access arguments' => array('manage projects'), 'page callback' => 'mymodule_projects_add', 'type' => MENU_NORMAL_ITEM, 'weight' => 1, ); return $items; } This will add a Projects section to the Administration area with an Add project sub section. All good. The behavior I want is that my client can only see the Projects section when they log in. I've accomplished this by ticking the "manage projects" permission for authenticated users. Now to give my client actual access to the Administration area I also need to tick "access administration pages" under the "system module" in the users permissions section. This works great, when I log in as my client I can only see the Projects section in the Administration area. There is one thing though, I my Navigation menu shown in the left column I can see the following items: - Administer - Projects - Content management - Site building - Site configuration - User management I was expecting only the see Administer and Projects, not the other ones. When I click e.g. Content Management I get a Content Management titled page with no options. Same for Site Building, Site Configuration and User Management. What's really odd is that Reports is not being shown which is also a top level Administration section. Why are these other items, besides my Projects section, being shown and how can I make them stop from appearing if I'm not logged in as administrator?

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  • Mistake in display and insert methods (double-ended queue)

    - by MANAL
    1) My problem when i make remove from right or left program will be remove true but when i call diplay method the content wrong like this I insert 12 43 65 23 and when make remove from left program will remove 12 but when call display method show like this 12 43 65 and when make remove from right program will remove 23 but when call display method show like this 12 43 Why ?????? ); and when i try to make insert after remove write this Can not insert right because the queue is full . first remove right and then u can insert right where is the problem ?? Please Help me please 2) My code FIRST CLASS class dqueue { private int fullsize; //number of all cells private int item_num; // number of busy cells only private int front,rear; public int j; private double [] dqarr; //========================================== public dqueue(int s) //constructor { fullsize = s; front = 0; rear = -1; item_num = 0; dqarr = new double[fullsize]; } //========================================== public void insert(double data) { if (rear == fullsize-1) rear = -1; rear++; dqarr[rear] = data; item_num++; } public double removeLeft() // take item from front of queue { double temp = dqarr[front++]; // get value and incr front if(front == fullsize) front = 0; item_num --; // one less item return temp; } public double removeRight() // take item from rear of queue { double temp = dqarr[rear--]; // get value and decr rear if(rear == -1) // rear = item_num -1; item_num --; // one less item return temp; } //========================================= public void display () //display items { for (int j=0;j<item_num;j++) // for every element System.out.print(dqarr[j] +" " ); // display it System.out.println(""); } //========================================= public int size() //number of items in queue { return item_num; } //========================================== public boolean isEmpty() // true if queue is empty { return (item_num ==0); } } SECOND CLASS import java.util.Scanner; class dqueuetest { public static void main(String[] args) { Scanner input = new Scanner(System.in); System.out.println(" ***** Welcome here***** "); System.out.println(" ***** Mind Of Programming Group***** "); System.out.println(" _____________________________________________ "); System.out.println("enter size of your dqueue"); int size = input.nextInt(); dqueue mydq = new dqueue(size); System.out.println(""); System.out.println("enter your itemes"); //===================================== for(int i = 0;i<=size-1;i++) { System.out.printf("item %d:",i+1); double item = input.nextDouble(); mydq.insert(item); System.out.println(""); } //===================================== int queue =size ; int c = 0 ; while (c != 6) { System.out.println(""); System.out.println("************************************************"); System.out.println(" MAIN MENUE"); System.out.println("1- INSERT RIGHT "); System.out.println("2- REMOVE LEFT"); System.out.println("3- REMOVE RIGHT"); System.out.println("4- DISPLAY"); System.out.println("5- SIZE"); System.out.println("6- EXIT"); System.out.println("************************************************"); System.out.println("choose your operation by number(1-6)"); c = input.nextInt(); switch (c) { case 1: if (queue == size) System.out.print("Can not insert right because the queue is full . first remove right and then u can insert right "); else { System.out.print("enter your item: "); double item = input.nextDouble(); mydq.insert(item);} break; case 2: System.out.println("REMOVE FROM REAR :"); if( !mydq.isEmpty() ) { double item = mydq.removeLeft(); System.out.print(item + "\t"); } // end while System.out.println(""); mydq.display(); break; case 3: System.out.println("REMOVE FROM FRONT :"); if( !mydq.isEmpty() ) { double item = mydq.removeRight(); System.out.print(item + "\t"); } // end while System.out.println(""); mydq.display(); break; case 4: System.out.println("The items in Queue are :"); mydq.display(); break; case 5: System.out.println("The Size of the Queue is :"+mydq.size()); break; case 6: System.out.println("Good Bye"); break; default: System.out.println("wrong chiose enter again"); } //end switch } //end while } // end main }//end class

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  • How can I change this method to get rid of the warning without anything changing?

    - by user3591323
    So this question:Warning-used as the name of the previous parameter rather than as part of the selector answers part of my problem, but I really don't want anything to change inside this method and I'm a bit confused on how this works. Here's the whole method: -(void) SetRightWrong:(sqzWord *)word: (int) rightWrong { if (self.mastered==nil) { self.mastered = [[NSMutableArray alloc]initWithCapacity:10]; } //if right change number right if (rightWrong == 1) { word.numberCorrect++; //if 3 right move to masterd list [self.onDeck removeObject:word]; if(word.numberCorrect >= 3 ) { [self.mastered addObject:word]; } else { //if not 3 right move to end of ondeck [self.onDeck addObject:word]; } } else if(rightWrong == 0) { //if wrong remove one from number right unless 0 NSUInteger i; i=[self.onDeck indexOfObject:word]; word = [self.onDeck objectAtIndex:i]; if (word.numberCorrect >0) { word.numberCorrect--; } } } The warning I am getting is: 'word' used as the name of the previous parameter than as part of the selector.

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  • Mistake in dispaly and insert method (double - ended queue)

    - by MANAL
    1) My problem when i make remove from right or left program will be remove true but when i call diplay method the content wrong like this I insert 12 43 65 23 and when make remove from left program will remove 12 but when call display method show like this 12 43 65 and when make remove from right program will remove 23 but when call display method show like this 12 43 Why ?????? ); and when i try to make insert after remove write this Can not insert right because the queue is full . first remove right and then u can insert right where is the problem ?? Please Help me please 2) My code FIRST CLASS class dqueue { private int fullsize; //number of all cells private int item_num; // number of busy cells only private int front,rear; public int j; private double [] dqarr; //========================================== public dqueue(int s) //constructor { fullsize = s; front = 0; rear = -1; item_num = 0; dqarr = new double[fullsize]; } //========================================== public void insert(double data) { if (rear == fullsize-1) rear = -1; rear++; dqarr[rear] = data; item_num++; } public double removeLeft() // take item from front of queue { double temp = dqarr[front++]; // get value and incr front if(front == fullsize) front = 0; item_num --; // one less item return temp; } public double removeRight() // take item from rear of queue { double temp = dqarr[rear--]; // get value and decr rear if(rear == -1) // rear = item_num -1; item_num --; // one less item return temp; } //========================================= public void display () //display items { for (int j=0;j //========================================= public int size() //number of items in queue { return item_num; } //========================================== public boolean isEmpty() // true if queue is empty { return (item_num ==0); } } SECOND CLASS import java.util.Scanner; class dqueuetest { public static void main(String[] args) { Scanner input = new Scanner(System.in); System.out.println(" Welcome here** "); System.out.println(" * Mind Of Programming Group*** "); System.out.println(" _________________________ "); System.out.println("enter size of your dqueue"); int size = input.nextInt(); dqueue mydq = new dqueue(size); System.out.println(""); System.out.println("enter your itemes"); //===================================== for(int i = 0;i<=size-1;i++) { System.out.printf("item %d:",i+1); double item = input.nextDouble(); mydq.insert(item); System.out.println(""); } //===================================== int queue =size ; int c = 0 ; while (c != 6) { System.out.println(""); System.out.println("**************************"); System.out.println(" MAIN MENUE"); System.out.println("1- INSERT RIGHT "); System.out.println("2- REMOVE LEFT"); System.out.println("3- REMOVE RIGHT"); System.out.println("4- DISPLAY"); System.out.println("5- SIZE"); System.out.println("6- EXIT"); System.out.println("**************************"); System.out.println("choose your operation by number(1-6)"); c = input.nextInt(); switch (c) { case 1: if (queue == size) System.out.print("Can not insert right because the queue is full . first remove right and then u can insert right "); else { System.out.print("enter your item: "); double item = input.nextDouble(); mydq.insert(item);} break; case 2: System.out.println("REMOVE FROM REAR :"); if( !mydq.isEmpty() ) { double item = mydq.removeLeft(); System.out.print(item + "\t"); } // end while System.out.println(""); mydq.display(); break; case 3: System.out.println("REMOVE FROM FRONT :"); if( !mydq.isEmpty() ) { double item = mydq.removeRight(); System.out.print(item + "\t"); } // end while System.out.println(""); mydq.display(); break; case 4: System.out.println("The items in Queue are :"); mydq.display(); break; case 5: System.out.println("The Size of the Queue is :"+mydq.size()); break; case 6: System.out.println("Good Bye"); break; default: System.out.println("wrong chiose enter again"); } //end switch } //end while } // end main }//end class

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