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  • Arithmetic Operations in Arrays with PHP

    - by Ajith
    I have two arrays $data1 = array() $data1 = array( '10','22','30') and also another array carries $data2 = array() $data2 = array( '2','11','3'); I need to divide these two arrays(ie,$data1/$data2) and store value to $data3[]. I need to get it as follows $data3[] = array('5','2','10') If someone knows of an easy way to do this would be of great help. Thanks

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  • Is it safe to make GL calls with multiple threads?

    - by user146780
    I was wondering if it was safe to make GL calls with multiple threads. Basically I'm using a GLUtesselator and was wondering if I could divide the objects to draw into 4 and assign a thread to each one. I'm just wondering if this would cause trouble since the tesselator uses callback functions. Can 2 threads run the same callback at the same time as long as that callback does not access ant global variables? Are there also other ways I could optimize OpenGL drawing using multithreading? Thanks

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  • matrices&searching [closed]

    - by gcc
    question 1) between different characters&real numbers , finding specific one how could i do question 2) myfriend asked me a good question : can we divide two matrices to each other // in math , we havenot learned but maybe someone knows where we find the answer

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  • C problem, left of '->' must point to class/struct/union/generic type ??

    - by Patrick
    Hello! Trying to understand why this doesn't work. I keep getting the following errors: left of '-nextNode' must point to class/struct/union/generic type (Also all the lines with a - in the function new_math_struct) Header file #ifndef MSTRUCT_H #define MSTRUCT_H #define PLUS 0 #define MINUS 1 #define DIVIDE 2 #define MULTIPLY 3 #define NUMBER 4 typedef struct math_struct { int type_of_value; int value; int sum; int is_used; struct math_struct* nextNode; } ; typedef struct math_struct* math_struct_ptr; #endif C file int get_input(math_struct_ptr* startNode) { /* character, input by the user */ char input_ch; char* input_ptr; math_struct_ptr* ptr; math_struct_ptr* previousNode; input_ptr = &input_ch; previousNode = startNode; /* as long as input is not ok */ while (1) { input_ch = get_input_character(); if (input_ch == ',') // Carrage return return 1; else if (input_ch == '.') // Illegal character return 0; if (input_ch == '+') ptr = new_math_struct(PLUS, 0); else if (input_ch == '-') ptr = new_math_struct(MINUS, 0); else if (input_ch == '/') ptr = new_math_struct(DIVIDE, 0); else if (input_ch == '*') ptr = new_math_struct(MULTIPLY, 0); else ptr = new_math_struct(NUMBER, atoi(input_ptr)); if (startNode == NULL) { startNode = previousNode = ptr; } else { previousNode->nextNode = ptr; previousNode = ptr; } } return 0; } math_struct_ptr* new_math_struct(int symbol, int value) { math_struct_ptr* ptr; ptr = (math_struct_ptr*)malloc(sizeof(math_struct_ptr)); ptr->type_of_value = symbol; ptr->value = value; ptr->sum = 0; ptr->is_used = 0; return ptr; } char get_input_character() { /* character, input by the user */ char input_ch; /* get the character */ scanf("%c", &input_ch); if (input_ch == '+' || input_ch == '-' || input_ch == '*' || input_ch == '/' || input_ch == ')') return input_ch; // A special character else if (input_ch == '\n') return ','; // A carrage return else if (input_ch < '0' || input_ch > '9') return '.'; // Not a number else return input_ch; // Number } The header for the C file just contains a reference to the struct header and the definitions of the functions. Language C.

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  • Returning the size of available virtual memory at run-time in C++

    - by Greenhouse Gases
    In C++ is there a predefined library function that will return the size of RAM currently available on a computer a program is being run on, at run-time? For instance, if an object is 4bytes, then can we divide the available virtual memory by 4 bytes to give an estimate of how many more objects could be stored by the program safely? I have used the sizeof() function to return the size of objects within my program. Thanks

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  • Basic string and value functions in Objective-C for iPhone

    - by David Maitland
    I have very little programming knowledge; only a fair bit in Visual Basic. How do I take a value from a text field, then do some simple math such as divide the value by two, then display it back to the user in the same field? In Visual Basic you could just do txtBoxOne.text = txtBoxOne.text / 2 I understand this question is more than one question and is very basic stuff, but I need to get my head out of Visual Basic and into where I should be :)

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  • Exceptions handling in SQL?

    - by Vineet
    Is there any way to handle exceptions in sql(ORACLE 9i)? Since I was trying to divide values of a column that contains both numbers and literals ,I need to fetch out only numbers from it ,as if it divisible by any number then its number else if contains literals it would not get divided it will generate error. how to handle those errors? Please suggest!!

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  • Safe division function

    - by bugspy.net
    I would like to define some kind of safe division (and modulo) function, one that would return some predefined value when attempting to divide by zero. I don't want to throw exceptions, just to return some "reasonable" value (1? 0?) and continue the program flow. Obviously there is no correct return value, but I wonder if there is some standard or known approach to this

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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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  • 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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  • How to follow object on CatmullRomSplines at constant speed (e.g. train and train carriage)?

    - by Simon
    I have a CatmullRomSpline, and using the very good example at https://github.com/libgdx/libgdx/wiki/Path-interface-%26-Splines I have my object moving at an even pace over the spline. Using a simple train and carriage example, I now want to have the carriage follow the train at the same speed as the train (not jolting along as it does with my code below). This leads into my main questions: How can I make the carriage have the same constant speed as the train and make it non jerky (it has something to do with the derivative I think, I don't understand how that part works)? Why do I need to divide by the line length to convert to metres per second, and is that correct? It wasn't done in the linked examples? I have used the example I linked to above, and modified for my specific example: private void process(CatmullRomSpline catmullRomSpline) { // Render path with precision of 1000 points renderPath(catmullRomSpline, 1000); float length = catmullRomSpline.approxLength(catmullRomSpline.spanCount * 1000); // Render the "train" Vector2 trainDerivative = new Vector2(); Vector2 trainLocation = new Vector2(); catmullRomSpline.derivativeAt(trainDerivative, current); // For some reason need to divide by length to convert from pixel speed to metres per second but I do not // really understand why I need it, it wasn't done in the examples??????? current += (Gdx.graphics.getDeltaTime() * speed / length) / trainDerivative.len(); catmullRomSpline.valueAt(trainLocation, current); renderCircleAtLocation(trainLocation); if (current >= 1) { current -= 1; } // Render the "carriage" Vector2 carriageLocation = new Vector2(); float carriagePercentageCovered = (((current * length) - 1f) / length); // I would like it to follow at 1 metre behind carriagePercentageCovered = Math.max(carriagePercentageCovered, 0); catmullRomSpline.valueAt(carriageLocation, carriagePercentageCovered); renderCircleAtLocation(carriageLocation); } private void renderPath(CatmullRomSpline catmullRomSpline, int k) { // catMulPoints would normally be cached when initialising, but for sake of example... Vector2[] catMulPoints = new Vector2[k]; for (int i = 0; i < k; ++i) { catMulPoints[i] = new Vector2(); catmullRomSpline.valueAt(catMulPoints[i], ((float) i) / ((float) k - 1)); } SHAPE_RENDERER.begin(ShapeRenderer.ShapeType.Line); SHAPE_RENDERER.setColor(Color.NAVY); for (int i = 0; i < k - 1; ++i) { SHAPE_RENDERER.line((Vector2) catMulPoints[i], (Vector2) catMulPoints[i + 1]); } SHAPE_RENDERER.end(); } private void renderCircleAtLocation(Vector2 location) { SHAPE_RENDERER.begin(ShapeRenderer.ShapeType.Filled); SHAPE_RENDERER.setColor(Color.YELLOW); SHAPE_RENDERER.circle(location.x, location.y, .5f); SHAPE_RENDERER.end(); } To create a decent sized CatmullRomSpline for testing this out: Vector2[] controlPoints = makeControlPointsArray(); CatmullRomSpline myCatmull = new CatmullRomSpline(controlPoints, false); .... private Vector2[] makeControlPointsArray() { Vector2[] pointsArray = new Vector2[78]; pointsArray[0] = new Vector2(1.681817f, 10.379999f); pointsArray[1] = new Vector2(2.045455f, 10.379999f); pointsArray[2] = new Vector2(2.663636f, 10.479999f); pointsArray[3] = new Vector2(3.027272f, 10.700000f); pointsArray[4] = new Vector2(3.663636f, 10.939999f); pointsArray[5] = new Vector2(4.245455f, 10.899999f); pointsArray[6] = new Vector2(4.736363f, 10.720000f); pointsArray[7] = new Vector2(4.754545f, 10.339999f); pointsArray[8] = new Vector2(4.518181f, 9.860000f); pointsArray[9] = new Vector2(3.790908f, 9.340000f); pointsArray[10] = new Vector2(3.172727f, 8.739999f); pointsArray[11] = new Vector2(3.300000f, 8.340000f); pointsArray[12] = new Vector2(3.700000f, 8.159999f); pointsArray[13] = new Vector2(4.227272f, 8.520000f); pointsArray[14] = new Vector2(4.681818f, 8.819999f); pointsArray[15] = new Vector2(5.081817f, 9.200000f); pointsArray[16] = new Vector2(5.463636f, 9.460000f); pointsArray[17] = new Vector2(5.972727f, 9.300000f); pointsArray[18] = new Vector2(6.063636f, 8.780000f); pointsArray[19] = new Vector2(6.027272f, 8.259999f); pointsArray[20] = new Vector2(5.700000f, 7.739999f); pointsArray[21] = new Vector2(5.300000f, 7.440000f); pointsArray[22] = new Vector2(4.645454f, 7.179999f); pointsArray[23] = new Vector2(4.136363f, 6.940000f); pointsArray[24] = new Vector2(3.427272f, 6.720000f); pointsArray[25] = new Vector2(2.572727f, 6.559999f); pointsArray[26] = new Vector2(1.900000f, 7.100000f); pointsArray[27] = new Vector2(2.336362f, 7.440000f); pointsArray[28] = new Vector2(2.590908f, 7.940000f); pointsArray[29] = new Vector2(2.318181f, 8.500000f); pointsArray[30] = new Vector2(1.663636f, 8.599999f); pointsArray[31] = new Vector2(1.209090f, 8.299999f); pointsArray[32] = new Vector2(1.118181f, 7.700000f); pointsArray[33] = new Vector2(1.045455f, 6.880000f); pointsArray[34] = new Vector2(1.154545f, 6.100000f); pointsArray[35] = new Vector2(1.281817f, 5.580000f); pointsArray[36] = new Vector2(1.700000f, 5.320000f); pointsArray[37] = new Vector2(2.190908f, 5.199999f); pointsArray[38] = new Vector2(2.900000f, 5.100000f); pointsArray[39] = new Vector2(3.700000f, 5.100000f); pointsArray[40] = new Vector2(4.372727f, 5.220000f); pointsArray[41] = new Vector2(4.827272f, 5.220000f); pointsArray[42] = new Vector2(5.463636f, 5.160000f); pointsArray[43] = new Vector2(5.554545f, 4.700000f); pointsArray[44] = new Vector2(5.245453f, 4.340000f); pointsArray[45] = new Vector2(4.445455f, 4.280000f); pointsArray[46] = new Vector2(3.609091f, 4.260000f); pointsArray[47] = new Vector2(2.718181f, 4.160000f); pointsArray[48] = new Vector2(1.990908f, 4.140000f); pointsArray[49] = new Vector2(1.427272f, 3.980000f); pointsArray[50] = new Vector2(1.609090f, 3.580000f); pointsArray[51] = new Vector2(2.136363f, 3.440000f); pointsArray[52] = new Vector2(3.227272f, 3.280000f); pointsArray[53] = new Vector2(3.972727f, 3.340000f); pointsArray[54] = new Vector2(5.027272f, 3.360000f); pointsArray[55] = new Vector2(5.718181f, 3.460000f); pointsArray[56] = new Vector2(6.100000f, 4.240000f); pointsArray[57] = new Vector2(6.209091f, 4.500000f); pointsArray[58] = new Vector2(6.118181f, 5.320000f); pointsArray[59] = new Vector2(5.772727f, 5.920000f); pointsArray[60] = new Vector2(4.881817f, 6.140000f); pointsArray[61] = new Vector2(5.318181f, 6.580000f); pointsArray[62] = new Vector2(6.263636f, 7.020000f); pointsArray[63] = new Vector2(6.645453f, 7.420000f); pointsArray[64] = new Vector2(6.681817f, 8.179999f); pointsArray[65] = new Vector2(6.627272f, 9.080000f); pointsArray[66] = new Vector2(6.572727f, 9.699999f); pointsArray[67] = new Vector2(6.263636f, 10.820000f); pointsArray[68] = new Vector2(5.754546f, 11.479999f); pointsArray[69] = new Vector2(4.536363f, 11.599998f); pointsArray[70] = new Vector2(3.572727f, 11.700000f); pointsArray[71] = new Vector2(2.809090f, 11.660000f); pointsArray[72] = new Vector2(1.445455f, 11.559999f); pointsArray[73] = new Vector2(0.936363f, 11.280000f); pointsArray[74] = new Vector2(0.754545f, 10.879999f); pointsArray[75] = new Vector2(0.700000f, 9.939999f); pointsArray[76] = new Vector2(0.918181f, 9.620000f); pointsArray[77] = new Vector2(1.463636f, 9.600000f); return pointsArray; } Disclaimer: My math is very rusty, so please explain in lay mans terms....

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  • Ask the Readers: Do You Use the Command Line?

    - by Asian Angel
    Most people have heard of it but not everyone is familiar or comfortable with how to use this bastion of geekdom. This week we would like to know if you use the command line or not. The command line…the bastion of ultimate geekery in many peoples’ eyes. You often hear people referring to doing things using the command line, so there must be something to it, right? For some people using the command line is the best, most efficient, and easiest way to do things on their systems. These are the people that many of us wish we were like. Next you have those who are proficient at using the command line but do not rely on it for everything they do on their systems. Then there are people who know how to perform some tasks or hacks using the command line but may not be as comfortable or knowledgeable as they wish to be using it. Moving on you find those who are interested in learning how to use the command line and just need a small push to get started.  Perhaps you feel too intimidated to learn it and just need the right opportunity to come along. And maybe you do not care one way or the other so long as you get done what you want to do on your system. Or you may prefer to simply use a graphical interface since that is quicker and easier for you (along with being familiar). You can find the whole range of people when it comes to using the command line… This week we would like to know if you use the command line or not. What command line category do you fit into? Power user? Casual usage? Totally lost? Let us know in the comments! How-To Geek Polls require Javascript. Please Click Here to View the Poll. Latest Features How-To Geek ETC The How-To Geek Holiday Gift Guide (Geeky Stuff We Like) LCD? LED? Plasma? The How-To Geek Guide to HDTV Technology The How-To Geek Guide to Learning Photoshop, Part 8: Filters Improve Digital Photography by Calibrating Your Monitor Our Favorite Tech: What We’re Thankful For at How-To Geek The How-To Geek Guide to Learning Photoshop, Part 7: Design and Typography Fun and Colorful Firefox Theme for Windows 7 Happy Snow Bears Theme for Chrome and Iron [Holiday] Download Full Command and Conquer: Tiberian Sun Game for Free Scorched Cometary Planet Wallpaper Quick Fix: Add the RSS Button Back to the Firefox Awesome Bar Dropbox Desktop Client 1.0.0 RC for Windows, Linux, and Mac Released

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  • Desktop Fun: Merry Christmas Fonts

    - by Asian Angel
    Christmas will soon be here and there are lots of cards, invitations, gift tags, photos, and more to prepare beforehand. To help you get ready we have gathered together a great collection of fun holiday fonts to help turn those ordinary looking holiday items into extraordinary looking ones. Note: To manage the fonts on your Windows 7, Vista, & XP systems see our article here. Oldchristmas Download Holly Download Christmas Flakes *includes two font types Download Frosty Download Kingthings Christmas Download Candy Time Download BodieMF Holly Download Snowfall Download Snowflake Letters Download Hultog Snowdrift Download AlphaShapes Xmas Trees Download Christmas Tree Download PF Wreath Download Snowy Caps Download PF Snowman *includes three font types Note: Shown in all capital letters here. Download BJF Holly Bells Download Christbaumkugeln Download Xmas Lights Download XmasDings *includes 62 individual characters Note: This group represents A – Z in all capital letters. Note: This group represents A – Z in all lower case letters. Note: This group represents the numbers 0 – 9. Download WWFlakes *includes 62 individual characters Note: This group represents A – Z in all capital letters. Note: This group represents A – Z in all lower case letters. Note: This group represents the numbers 0 – 9. Download For Christmas Card creating fun and a great way to use your new fonts see our MS Word Christmas Card project series here. Design and Print Your Own Christmas Cards in MS Word, Part 1 Design and Print Your Own Christmas Cards in MS Word, Part 2: How to Print Want more great ways to customize your computer? Then be certain to look through our Desktop Fun section. Latest Features How-To Geek ETC The How-To Geek Holiday Gift Guide (Geeky Stuff We Like) LCD? LED? Plasma? The How-To Geek Guide to HDTV Technology The How-To Geek Guide to Learning Photoshop, Part 8: Filters Improve Digital Photography by Calibrating Your Monitor Our Favorite Tech: What We’re Thankful For at How-To Geek The How-To Geek Guide to Learning Photoshop, Part 7: Design and Typography Happy Snow Bears Theme for Chrome and Iron [Holiday] Download Full Command and Conquer: Tiberian Sun Game for Free Scorched Cometary Planet Wallpaper Quick Fix: Add the RSS Button Back to the Firefox Awesome Bar Dropbox Desktop Client 1.0.0 RC for Windows, Linux, and Mac Released Hang in There Scrat! – Ice Age Wallpaper

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  • Improving Partitioned Table Join Performance

    - by Paul White
    The query optimizer does not always choose an optimal strategy when joining partitioned tables. This post looks at an example, showing how a manual rewrite of the query can almost double performance, while reducing the memory grant to almost nothing. Test Data The two tables in this example use a common partitioning partition scheme. The partition function uses 41 equal-size partitions: CREATE PARTITION FUNCTION PFT (integer) AS RANGE RIGHT FOR VALUES ( 125000, 250000, 375000, 500000, 625000, 750000, 875000, 1000000, 1125000, 1250000, 1375000, 1500000, 1625000, 1750000, 1875000, 2000000, 2125000, 2250000, 2375000, 2500000, 2625000, 2750000, 2875000, 3000000, 3125000, 3250000, 3375000, 3500000, 3625000, 3750000, 3875000, 4000000, 4125000, 4250000, 4375000, 4500000, 4625000, 4750000, 4875000, 5000000 ); GO CREATE PARTITION SCHEME PST AS PARTITION PFT ALL TO ([PRIMARY]); There two tables are: CREATE TABLE dbo.T1 ( TID integer NOT NULL IDENTITY(0,1), Column1 integer NOT NULL, Padding binary(100) NOT NULL DEFAULT 0x,   CONSTRAINT PK_T1 PRIMARY KEY CLUSTERED (TID) ON PST (TID) );   CREATE TABLE dbo.T2 ( TID integer NOT NULL, Column1 integer NOT NULL, Padding binary(100) NOT NULL DEFAULT 0x,   CONSTRAINT PK_T2 PRIMARY KEY CLUSTERED (TID, Column1) ON PST (TID) ); The next script loads 5 million rows into T1 with a pseudo-random value between 1 and 5 for Column1. The table is partitioned on the IDENTITY column TID: INSERT dbo.T1 WITH (TABLOCKX) (Column1) SELECT (ABS(CHECKSUM(NEWID())) % 5) + 1 FROM dbo.Numbers AS N WHERE n BETWEEN 1 AND 5000000; In case you don’t already have an auxiliary table of numbers lying around, here’s a script to create one with 10 million rows: CREATE TABLE dbo.Numbers (n bigint PRIMARY KEY);   WITH L0 AS(SELECT 1 AS c UNION ALL SELECT 1), L1 AS(SELECT 1 AS c FROM L0 AS A CROSS JOIN L0 AS B), L2 AS(SELECT 1 AS c FROM L1 AS A CROSS JOIN L1 AS B), L3 AS(SELECT 1 AS c FROM L2 AS A CROSS JOIN L2 AS B), L4 AS(SELECT 1 AS c FROM L3 AS A CROSS JOIN L3 AS B), L5 AS(SELECT 1 AS c FROM L4 AS A CROSS JOIN L4 AS B), Nums AS(SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) AS n FROM L5) INSERT dbo.Numbers WITH (TABLOCKX) SELECT TOP (10000000) n FROM Nums ORDER BY n OPTION (MAXDOP 1); Table T1 contains data like this: Next we load data into table T2. The relationship between the two tables is that table 2 contains ‘n’ rows for each row in table 1, where ‘n’ is determined by the value in Column1 of table T1. There is nothing particularly special about the data or distribution, by the way. INSERT dbo.T2 WITH (TABLOCKX) (TID, Column1) SELECT T.TID, N.n FROM dbo.T1 AS T JOIN dbo.Numbers AS N ON N.n >= 1 AND N.n <= T.Column1; Table T2 ends up containing about 15 million rows: The primary key for table T2 is a combination of TID and Column1. The data is partitioned according to the value in column TID alone. Partition Distribution The following query shows the number of rows in each partition of table T1: SELECT PartitionID = CA1.P, NumRows = COUNT_BIG(*) FROM dbo.T1 AS T CROSS APPLY (VALUES ($PARTITION.PFT(TID))) AS CA1 (P) GROUP BY CA1.P ORDER BY CA1.P; There are 40 partitions containing 125,000 rows (40 * 125k = 5m rows). The rightmost partition remains empty. The next query shows the distribution for table 2: SELECT PartitionID = CA1.P, NumRows = COUNT_BIG(*) FROM dbo.T2 AS T CROSS APPLY (VALUES ($PARTITION.PFT(TID))) AS CA1 (P) GROUP BY CA1.P ORDER BY CA1.P; There are roughly 375,000 rows in each partition (the rightmost partition is also empty): Ok, that’s the test data done. Test Query and Execution Plan The task is to count the rows resulting from joining tables 1 and 2 on the TID column: SET STATISTICS IO ON; DECLARE @s datetime2 = SYSUTCDATETIME();   SELECT COUNT_BIG(*) FROM dbo.T1 AS T1 JOIN dbo.T2 AS T2 ON T2.TID = T1.TID;   SELECT DATEDIFF(Millisecond, @s, SYSUTCDATETIME()); SET STATISTICS IO OFF; The optimizer chooses a plan using parallel hash join, and partial aggregation: The Plan Explorer plan tree view shows accurate cardinality estimates and an even distribution of rows across threads (click to enlarge the image): With a warm data cache, the STATISTICS IO output shows that no physical I/O was needed, and all 41 partitions were touched: Running the query without actual execution plan or STATISTICS IO information for maximum performance, the query returns in around 2600ms. Execution Plan Analysis The first step toward improving on the execution plan produced by the query optimizer is to understand how it works, at least in outline. The two parallel Clustered Index Scans use multiple threads to read rows from tables T1 and T2. Parallel scan uses a demand-based scheme where threads are given page(s) to scan from the table as needed. This arrangement has certain important advantages, but does result in an unpredictable distribution of rows amongst threads. The point is that multiple threads cooperate to scan the whole table, but it is impossible to predict which rows end up on which threads. For correct results from the parallel hash join, the execution plan has to ensure that rows from T1 and T2 that might join are processed on the same thread. For example, if a row from T1 with join key value ‘1234’ is placed in thread 5’s hash table, the execution plan must guarantee that any rows from T2 that also have join key value ‘1234’ probe thread 5’s hash table for matches. The way this guarantee is enforced in this parallel hash join plan is by repartitioning rows to threads after each parallel scan. The two repartitioning exchanges route rows to threads using a hash function over the hash join keys. The two repartitioning exchanges use the same hash function so rows from T1 and T2 with the same join key must end up on the same hash join thread. Expensive Exchanges This business of repartitioning rows between threads can be very expensive, especially if a large number of rows is involved. The execution plan selected by the optimizer moves 5 million rows through one repartitioning exchange and around 15 million across the other. As a first step toward removing these exchanges, consider the execution plan selected by the optimizer if we join just one partition from each table, disallowing parallelism: SELECT COUNT_BIG(*) FROM dbo.T1 AS T1 JOIN dbo.T2 AS T2 ON T2.TID = T1.TID WHERE $PARTITION.PFT(T1.TID) = 1 AND $PARTITION.PFT(T2.TID) = 1 OPTION (MAXDOP 1); The optimizer has chosen a (one-to-many) merge join instead of a hash join. The single-partition query completes in around 100ms. If everything scaled linearly, we would expect that extending this strategy to all 40 populated partitions would result in an execution time around 4000ms. Using parallelism could reduce that further, perhaps to be competitive with the parallel hash join chosen by the optimizer. This raises a question. If the most efficient way to join one partition from each of the tables is to use a merge join, why does the optimizer not choose a merge join for the full query? Forcing a Merge Join Let’s force the optimizer to use a merge join on the test query using a hint: SELECT COUNT_BIG(*) FROM dbo.T1 AS T1 JOIN dbo.T2 AS T2 ON T2.TID = T1.TID OPTION (MERGE JOIN); This is the execution plan selected by the optimizer: This plan results in the same number of logical reads reported previously, but instead of 2600ms the query takes 5000ms. The natural explanation for this drop in performance is that the merge join plan is only using a single thread, whereas the parallel hash join plan could use multiple threads. Parallel Merge Join We can get a parallel merge join plan using the same query hint as before, and adding trace flag 8649: SELECT COUNT_BIG(*) FROM dbo.T1 AS T1 JOIN dbo.T2 AS T2 ON T2.TID = T1.TID OPTION (MERGE JOIN, QUERYTRACEON 8649); The execution plan is: This looks promising. It uses a similar strategy to distribute work across threads as seen for the parallel hash join. In practice though, performance is disappointing. On a typical run, the parallel merge plan runs for around 8400ms; slower than the single-threaded merge join plan (5000ms) and much worse than the 2600ms for the parallel hash join. We seem to be going backwards! The logical reads for the parallel merge are still exactly the same as before, with no physical IOs. The cardinality estimates and thread distribution are also still very good (click to enlarge): A big clue to the reason for the poor performance is shown in the wait statistics (captured by Plan Explorer Pro): CXPACKET waits require careful interpretation, and are most often benign, but in this case excessive waiting occurs at the repartitioning exchanges. Unlike the parallel hash join, the repartitioning exchanges in this plan are order-preserving ‘merging’ exchanges (because merge join requires ordered inputs): Parallelism works best when threads can just grab any available unit of work and get on with processing it. Preserving order introduces inter-thread dependencies that can easily lead to significant waits occurring. In extreme cases, these dependencies can result in an intra-query deadlock, though the details of that will have to wait for another time to explore in detail. The potential for waits and deadlocks leads the query optimizer to cost parallel merge join relatively highly, especially as the degree of parallelism (DOP) increases. This high costing resulted in the optimizer choosing a serial merge join rather than parallel in this case. The test results certainly confirm its reasoning. Collocated Joins In SQL Server 2008 and later, the optimizer has another available strategy when joining tables that share a common partition scheme. This strategy is a collocated join, also known as as a per-partition join. It can be applied in both serial and parallel execution plans, though it is limited to 2-way joins in the current optimizer. Whether the optimizer chooses a collocated join or not depends on cost estimation. The primary benefits of a collocated join are that it eliminates an exchange and requires less memory, as we will see next. Costing and Plan Selection The query optimizer did consider a collocated join for our original query, but it was rejected on cost grounds. The parallel hash join with repartitioning exchanges appeared to be a cheaper option. There is no query hint to force a collocated join, so we have to mess with the costing framework to produce one for our test query. Pretending that IOs cost 50 times more than usual is enough to convince the optimizer to use collocated join with our test query: -- Pretend IOs are 50x cost temporarily DBCC SETIOWEIGHT(50);   -- Co-located hash join SELECT COUNT_BIG(*) FROM dbo.T1 AS T1 JOIN dbo.T2 AS T2 ON T2.TID = T1.TID OPTION (RECOMPILE);   -- Reset IO costing DBCC SETIOWEIGHT(1); Collocated Join Plan The estimated execution plan for the collocated join is: The Constant Scan contains one row for each partition of the shared partitioning scheme, from 1 to 41. The hash repartitioning exchanges seen previously are replaced by a single Distribute Streams exchange using Demand partitioning. Demand partitioning means that the next partition id is given to the next parallel thread that asks for one. My test machine has eight logical processors, and all are available for SQL Server to use. As a result, there are eight threads in the single parallel branch in this plan, each processing one partition from each table at a time. Once a thread finishes processing a partition, it grabs a new partition number from the Distribute Streams exchange…and so on until all partitions have been processed. It is important to understand that the parallel scans in this plan are different from the parallel hash join plan. Although the scans have the same parallelism icon, tables T1 and T2 are not being co-operatively scanned by multiple threads in the same way. Each thread reads a single partition of T1 and performs a hash match join with the same partition from table T2. The properties of the two Clustered Index Scans show a Seek Predicate (unusual for a scan!) limiting the rows to a single partition: The crucial point is that the join between T1 and T2 is on TID, and TID is the partitioning column for both tables. A thread that processes partition ‘n’ is guaranteed to see all rows that can possibly join on TID for that partition. In addition, no other thread will see rows from that partition, so this removes the need for repartitioning exchanges. CPU and Memory Efficiency Improvements The collocated join has removed two expensive repartitioning exchanges and added a single exchange processing 41 rows (one for each partition id). Remember, the parallel hash join plan exchanges had to process 5 million and 15 million rows. The amount of processor time spent on exchanges will be much lower in the collocated join plan. In addition, the collocated join plan has a maximum of 8 threads processing single partitions at any one time. The 41 partitions will all be processed eventually, but a new partition is not started until a thread asks for it. Threads can reuse hash table memory for the new partition. The parallel hash join plan also had 8 hash tables, but with all 5,000,000 build rows loaded at the same time. The collocated plan needs memory for only 8 * 125,000 = 1,000,000 rows at any one time. Collocated Hash Join Performance The collated join plan has disappointing performance in this case. The query runs for around 25,300ms despite the same IO statistics as usual. This is much the worst result so far, so what went wrong? It turns out that cardinality estimation for the single partition scans of table T1 is slightly low. The properties of the Clustered Index Scan of T1 (graphic immediately above) show the estimation was for 121,951 rows. This is a small shortfall compared with the 125,000 rows actually encountered, but it was enough to cause the hash join to spill to physical tempdb: A level 1 spill doesn’t sound too bad, until you realize that the spill to tempdb probably occurs for each of the 41 partitions. As a side note, the cardinality estimation error is a little surprising because the system tables accurately show there are 125,000 rows in every partition of T1. Unfortunately, the optimizer uses regular column and index statistics to derive cardinality estimates here rather than system table information (e.g. sys.partitions). Collocated Merge Join We will never know how well the collocated parallel hash join plan might have worked without the cardinality estimation error (and the resulting 41 spills to tempdb) but we do know: Merge join does not require a memory grant; and Merge join was the optimizer’s preferred join option for a single partition join Putting this all together, what we would really like to see is the same collocated join strategy, but using merge join instead of hash join. Unfortunately, the current query optimizer cannot produce a collocated merge join; it only knows how to do collocated hash join. So where does this leave us? CROSS APPLY sys.partitions We can try to write our own collocated join query. We can use sys.partitions to find the partition numbers, and CROSS APPLY to get a count per partition, with a final step to sum the partial counts. The following query implements this idea: SELECT row_count = SUM(Subtotals.cnt) FROM ( -- Partition numbers SELECT p.partition_number FROM sys.partitions AS p WHERE p.[object_id] = OBJECT_ID(N'T1', N'U') AND p.index_id = 1 ) AS P CROSS APPLY ( -- Count per collocated join SELECT cnt = COUNT_BIG(*) FROM dbo.T1 AS T1 JOIN dbo.T2 AS T2 ON T2.TID = T1.TID WHERE $PARTITION.PFT(T1.TID) = p.partition_number AND $PARTITION.PFT(T2.TID) = p.partition_number ) AS SubTotals; The estimated plan is: The cardinality estimates aren’t all that good here, especially the estimate for the scan of the system table underlying the sys.partitions view. Nevertheless, the plan shape is heading toward where we would like to be. Each partition number from the system table results in a per-partition scan of T1 and T2, a one-to-many Merge Join, and a Stream Aggregate to compute the partial counts. The final Stream Aggregate just sums the partial counts. Execution time for this query is around 3,500ms, with the same IO statistics as always. This compares favourably with 5,000ms for the serial plan produced by the optimizer with the OPTION (MERGE JOIN) hint. This is another case of the sum of the parts being less than the whole – summing 41 partial counts from 41 single-partition merge joins is faster than a single merge join and count over all partitions. Even so, this single-threaded collocated merge join is not as quick as the original parallel hash join plan, which executed in 2,600ms. On the positive side, our collocated merge join uses only one logical processor and requires no memory grant. The parallel hash join plan used 16 threads and reserved 569 MB of memory:   Using a Temporary Table Our collocated merge join plan should benefit from parallelism. The reason parallelism is not being used is that the query references a system table. We can work around that by writing the partition numbers to a temporary table (or table variable): SET STATISTICS IO ON; DECLARE @s datetime2 = SYSUTCDATETIME();   CREATE TABLE #P ( partition_number integer PRIMARY KEY);   INSERT #P (partition_number) SELECT p.partition_number FROM sys.partitions AS p WHERE p.[object_id] = OBJECT_ID(N'T1', N'U') AND p.index_id = 1;   SELECT row_count = SUM(Subtotals.cnt) FROM #P AS p CROSS APPLY ( SELECT cnt = COUNT_BIG(*) FROM dbo.T1 AS T1 JOIN dbo.T2 AS T2 ON T2.TID = T1.TID WHERE $PARTITION.PFT(T1.TID) = p.partition_number AND $PARTITION.PFT(T2.TID) = p.partition_number ) AS SubTotals;   DROP TABLE #P;   SELECT DATEDIFF(Millisecond, @s, SYSUTCDATETIME()); SET STATISTICS IO OFF; Using the temporary table adds a few logical reads, but the overall execution time is still around 3500ms, indistinguishable from the same query without the temporary table. The problem is that the query optimizer still doesn’t choose a parallel plan for this query, though the removal of the system table reference means that it could if it chose to: In fact the optimizer did enter the parallel plan phase of query optimization (running search 1 for a second time): Unfortunately, the parallel plan found seemed to be more expensive than the serial plan. This is a crazy result, caused by the optimizer’s cost model not reducing operator CPU costs on the inner side of a nested loops join. Don’t get me started on that, we’ll be here all night. In this plan, everything expensive happens on the inner side of a nested loops join. Without a CPU cost reduction to compensate for the added cost of exchange operators, candidate parallel plans always look more expensive to the optimizer than the equivalent serial plan. Parallel Collocated Merge Join We can produce the desired parallel plan using trace flag 8649 again: SELECT row_count = SUM(Subtotals.cnt) FROM #P AS p CROSS APPLY ( SELECT cnt = COUNT_BIG(*) FROM dbo.T1 AS T1 JOIN dbo.T2 AS T2 ON T2.TID = T1.TID WHERE $PARTITION.PFT(T1.TID) = p.partition_number AND $PARTITION.PFT(T2.TID) = p.partition_number ) AS SubTotals OPTION (QUERYTRACEON 8649); The actual execution plan is: One difference between this plan and the collocated hash join plan is that a Repartition Streams exchange operator is used instead of Distribute Streams. The effect is similar, though not quite identical. The Repartition uses round-robin partitioning, meaning the next partition id is pushed to the next thread in sequence. The Distribute Streams exchange seen earlier used Demand partitioning, meaning the next partition id is pulled across the exchange by the next thread that is ready for more work. There are subtle performance implications for each partitioning option, but going into that would again take us too far off the main point of this post. Performance The important thing is the performance of this parallel collocated merge join – just 1350ms on a typical run. The list below shows all the alternatives from this post (all timings include creation, population, and deletion of the temporary table where appropriate) from quickest to slowest: Collocated parallel merge join: 1350ms Parallel hash join: 2600ms Collocated serial merge join: 3500ms Serial merge join: 5000ms Parallel merge join: 8400ms Collated parallel hash join: 25,300ms (hash spill per partition) The parallel collocated merge join requires no memory grant (aside from a paltry 1.2MB used for exchange buffers). This plan uses 16 threads at DOP 8; but 8 of those are (rather pointlessly) allocated to the parallel scan of the temporary table. These are minor concerns, but it turns out there is a way to address them if it bothers you. Parallel Collocated Merge Join with Demand Partitioning This final tweak replaces the temporary table with a hard-coded list of partition ids (dynamic SQL could be used to generate this query from sys.partitions): SELECT row_count = SUM(Subtotals.cnt) FROM ( VALUES (1),(2),(3),(4),(5),(6),(7),(8),(9),(10), (11),(12),(13),(14),(15),(16),(17),(18),(19),(20), (21),(22),(23),(24),(25),(26),(27),(28),(29),(30), (31),(32),(33),(34),(35),(36),(37),(38),(39),(40),(41) ) AS P (partition_number) CROSS APPLY ( SELECT cnt = COUNT_BIG(*) FROM dbo.T1 AS T1 JOIN dbo.T2 AS T2 ON T2.TID = T1.TID WHERE $PARTITION.PFT(T1.TID) = p.partition_number AND $PARTITION.PFT(T2.TID) = p.partition_number ) AS SubTotals OPTION (QUERYTRACEON 8649); The actual execution plan is: The parallel collocated hash join plan is reproduced below for comparison: The manual rewrite has another advantage that has not been mentioned so far: the partial counts (per partition) can be computed earlier than the partial counts (per thread) in the optimizer’s collocated join plan. The earlier aggregation is performed by the extra Stream Aggregate under the nested loops join. The performance of the parallel collocated merge join is unchanged at around 1350ms. Final Words It is a shame that the current query optimizer does not consider a collocated merge join (Connect item closed as Won’t Fix). The example used in this post showed an improvement in execution time from 2600ms to 1350ms using a modestly-sized data set and limited parallelism. In addition, the memory requirement for the query was almost completely eliminated  – down from 569MB to 1.2MB. The problem with the parallel hash join selected by the optimizer is that it attempts to process the full data set all at once (albeit using eight threads). It requires a large memory grant to hold all 5 million rows from table T1 across the eight hash tables, and does not take advantage of the divide-and-conquer opportunity offered by the common partitioning. The great thing about the collocated join strategies is that each parallel thread works on a single partition from both tables, reading rows, performing the join, and computing a per-partition subtotal, before moving on to a new partition. From a thread’s point of view… If you have trouble visualizing what is happening from just looking at the parallel collocated merge join execution plan, let’s look at it again, but from the point of view of just one thread operating between the two Parallelism (exchange) operators. Our thread picks up a single partition id from the Distribute Streams exchange, and starts a merge join using ordered rows from partition 1 of table T1 and partition 1 of table T2. By definition, this is all happening on a single thread. As rows join, they are added to a (per-partition) count in the Stream Aggregate immediately above the Merge Join. Eventually, either T1 (partition 1) or T2 (partition 1) runs out of rows and the merge join stops. The per-partition count from the aggregate passes on through the Nested Loops join to another Stream Aggregate, which is maintaining a per-thread subtotal. Our same thread now picks up a new partition id from the exchange (say it gets id 9 this time). The count in the per-partition aggregate is reset to zero, and the processing of partition 9 of both tables proceeds just as it did for partition 1, and on the same thread. Each thread picks up a single partition id and processes all the data for that partition, completely independently from other threads working on other partitions. One thread might eventually process partitions (1, 9, 17, 25, 33, 41) while another is concurrently processing partitions (2, 10, 18, 26, 34) and so on for the other six threads at DOP 8. The point is that all 8 threads can execute independently and concurrently, continuing to process new partitions until the wider job (of which the thread has no knowledge!) is done. This divide-and-conquer technique can be much more efficient than simply splitting the entire workload across eight threads all at once. Related Reading Understanding and Using Parallelism in SQL Server Parallel Execution Plans Suck © 2013 Paul White – All Rights Reserved Twitter: @SQL_Kiwi

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  • What Precalculus knowledge is required before learning Discrete Math Computer Science topics?

    - by Ein Doofus
    Below I've listed the chapters from a Precalculus book as well as the author recommended Computer Science chapters from a Discrete Mathematics book. Although these chapters are from two specific books on these subjects I believe the topics are generally the same between any Precalc or Discrete Math book. What Precalculus topics should one know before starting these Discrete Math Computer Science topics?: Discrete Mathematics CS Chapters 1.1 Propositional Logic 1.2 Propositional Equivalences 1.3 Predicates and Quantifiers 1.4 Nested Quantifiers 1.5 Rules of Inference 1.6 Introduction to Proofs 1.7 Proof Methods and Strategy 2.1 Sets 2.2 Set Operations 2.3 Functions 2.4 Sequences and Summations 3.1 Algorithms 3.2 The Growths of Functions 3.3 Complexity of Algorithms 3.4 The Integers and Division 3.5 Primes and Greatest Common Divisors 3.6 Integers and Algorithms 3.8 Matrices 4.1 Mathematical Induction 4.2 Strong Induction and Well-Ordering 4.3 Recursive Definitions and Structural Induction 4.4 Recursive Algorithms 4.5 Program Correctness 5.1 The Basics of Counting 5.2 The Pigeonhole Principle 5.3 Permutations and Combinations 5.6 Generating Permutations and Combinations 6.1 An Introduction to Discrete Probability 6.4 Expected Value and Variance 7.1 Recurrence Relations 7.3 Divide-and-Conquer Algorithms and Recurrence Relations 7.5 Inclusion-Exclusion 8.1 Relations and Their Properties 8.2 n-ary Relations and Their Applications 8.3 Representing Relations 8.5 Equivalence Relations 9.1 Graphs and Graph Models 9.2 Graph Terminology and Special Types of Graphs 9.3 Representing Graphs and Graph Isomorphism 9.4 Connectivity 9.5 Euler and Hamilton Ptahs 10.1 Introduction to Trees 10.2 Application of Trees 10.3 Tree Traversal 11.1 Boolean Functions 11.2 Representing Boolean Functions 11.3 Logic Gates 11.4 Minimization of Circuits 12.1 Language and Grammars 12.2 Finite-State Machines with Output 12.3 Finite-State Machines with No Output 12.4 Language Recognition 12.5 Turing Machines Precalculus Chapters R.1 The Real-Number System R.2 Integer Exponents, Scientific Notation, and Order of Operations R.3 Addition, Subtraction, and Multiplication of Polynomials R.4 Factoring R.5 Rational Expressions R.6 Radical Notation and Rational Exponents R.7 The Basics of Equation Solving 1.1 Functions, Graphs, Graphers 1.2 Linear Functions, Slope, and Applications 1.3 Modeling: Data Analysis, Curve Fitting, and Linear Regression 1.4 More on Functions 1.5 Symmetry and Transformations 1.6 Variation and Applications 1.7 Distance, Midpoints, and Circles 2.1 Zeros of Linear Functions and Models 2.2 The Complex Numbers 2.3 Zeros of Quadratic Functions and Models 2.4 Analyzing Graphs of Quadratic Functions 2.5 Modeling: Data Analysis, Curve Fitting, and Quadratic Regression 2.6 Zeros and More Equation Solving 2.7 Solving Inequalities 3.1 Polynomial Functions and Modeling 3.2 Polynomial Division; The Remainder and Factor Theorems 3.3 Theorems about Zeros of Polynomial Functions 3.4 Rational Functions 3.5 Polynomial and Rational Inequalities 4.1 Composite and Inverse Functions 4.2 Exponential Functions and Graphs 4.3 Logarithmic Functions and Graphs 4.4 Properties of Logarithmic Functions 4.5 Solving Exponential and Logarithmic Equations 4.6 Applications and Models: Growth and Decay 5.1 Systems of Equations in Two Variables 5.2 System of Equations in Three Variables 5.3 Matrices and Systems of Equations 5.4 Matrix Operations 5.5 Inverses of Matrices 5.6 System of Inequalities and Linear Programming 5.7 Partial Fractions 6.1 The Parabola 6.2 The Circle and Ellipse 6.3 The Hyperbola 6.4 Nonlinear Systems of Equations

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  • How can I diagnose/debug "maximum number of clients reached" X errors?

    - by jmtd
    Hi, I'm hitting a problem whereby X prevents processes from creating windows, uttering something like the following into ~/.xsession-errors: cannot open display: :0.0 Maximum number of clients reached Searching around there are lots of examples of people facing this problem, and sometimes people identify which program they are running is using up all the client slots. See e.g. LP 70872 (Firefox), LP 263211 (gnome-screensaver). For what it's worth, I run gnome-terminal, thunderbird, chromium-browser, empathy, tomboy and virtualbox nearly all the time, on top of the normal stuff you get with the GNOME desktop, and occasionally some other bits and pieces. However my question is not "which of my programs is causing this problem" but rather, how can one go about diagnosing this problem? In the above (and other) bugs, forum reports, etc., a number of tools are suggested: xlsclients - lists the client applications for the given display, but I don't think that corresponds to 'X clients' xrestop - a top-style X resources tool, one row per X client. Lots of '' clients, not shown in xlsclients output xwininfo -root -children lists X window objects From what I can gather, the problem might not be too many clients at all, but rather resources kept around in the X server for clients who have long-since detached. But it would also appear that you cannot (easily?) relate X resources back to their client. Can one effectively diagnose this issue once it has started to occur, or is a tedious divide-and-conquer approach for the apps I run the only approach open to me? Update Jan 2011: I think I have resolved this issue. For the benefit of anyone stumbling across this, nautilus and/or compiz or something in that chain of software was segfaulting due to a wallpaper I had. I had chosen an XML file as my wallpaper, which defined a rotating gallery of images. It was hand-made, but based on /usr/share/backgrounds/contest/background-1.xml or similar. Disabling the wallpaper and I have not had a crash since. I'm not marking this as answered yet, since the actual specific problem was not my question, but how to diagnose it was. Unfortunately this was mostly trial-and-error which sucks.

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  • Alright, I'm still stuck on this homework problem. C++

    - by Josh
    Okay, the past few days I have been trying to get some input on my programs. Well I decided to scrap them for the most part and try again. So once again, I'm in need of help. For the first program I'm trying to fix, it needs to show the sum of SEVEN numbers. Well, I'm trying to change is so that I don't need the mem[##] = ####. I just want the user to be able to input the numbers and the program run from there and go through my switch loop. And have some kind of display..saying like the sum is?.. Here's my code so far. #include <iostream> #include <iomanip> #include <ios> using namespace std; int main() { const int READ = 10; const int WRITE = 11; const int LOAD = 20; const int STORE = 21; const int ADD = 30; const int SUBTRACT = 31; const int DIVIDE = 32; const int MULTIPLY = 33; const int BRANCH = 40; const int BRANCHNEG = 41; const int BRANCHZERO = 42; const int HALT = 43; int mem[100] = {0}; //Making it 100, since simpletron contains a 100 word mem. int operation; //taking the rest of these variables straight out of the book seeing as how they were italisized. int operand; int accum = 0; // the special register is starting at 0 int counter; for ( counter=0; counter < 100; counter++) mem[counter] = 0; // This is for part a, it will take in positive variables in //a sent-controlled loop and compute + print their sum. Variables from example in text. mem[0] = 1009; mem[1] = 1109; mem[2] = 2010; mem[3] = 2111; mem[4] = 2011; mem[5] = 3100; mem[6] = 2113; mem[7] = 1113; mem[8] = 4300; counter = 0; //Makes the variable counter start at 0. while(true) { operand = mem[ counter ]%100; // Finds the op codes from the limit on the mem (100) operation = mem[ counter ]/100; //using a switch loop to set up the loops for the cases switch ( operation ){ case READ: //reads a variable into a word from loc. Enter in -1 to exit cout <<"\n Input a positive variable: "; cin >> mem[ operand ]; counter++; break; case WRITE: // takes a word from location cout << "\n\nThe content at location " << operand << " is " << mem[operand]; counter++; break; case LOAD:// loads accum = mem[ operand ];counter++; break; case STORE: //stores mem[ operand ] = accum;counter++; break; case ADD: //adds accum += mem[operand];counter++; break; case SUBTRACT: // subtracts accum-= mem[ operand ];counter++; break; case DIVIDE: //divides accum /=(mem[ operand ]);counter++; break; case MULTIPLY: // multiplies accum*= mem [ operand ];counter++; break; case BRANCH: // Branches to location counter = operand; break; case BRANCHNEG: //branches if acc. is < 0 if (accum < 0) counter = operand; else counter++; break; case BRANCHZERO: //branches if acc = 0 if (accum == 0) counter = operand; else counter++; break; case HALT: // Program ends break; } } return 0; } part B int main() { const int READ = 10; const int WRITE = 11; const int LOAD = 20; const int STORE = 21; const int ADD = 30; const int SUBTRACT = 31; const int DIVIDE = 32; const int MULTIPLY = 33; const int BRANCH = 40; const int BRANCHNEG = 41; const int BRANCHZERO = 41; const int HALT = 43; int mem[100] = {0}; int operation; int operand; int accum = 0; int pos = 0; int j; mem[22] = 7; // loop 7 times mem[25] = 1; // increment by 1 mem[00] = 4306; mem[01] = 2303; mem[02] = 3402; mem[03] = 6410; mem[04] = 3412; mem[05] = 2111; mem[06] = 2002; mem[07] = 2312; mem[08] = 4210; mem[09] = 2109; mem[10] = 4001; mem[11] = 2015; mem[12] = 3212; mem[13] = 2116; mem[14] = 1101; mem[15] = 1116; mem[16] = 4300; j = 0; while ( true ) { operand = memory[ j ]%100; // Finds the op codes from the limit on the memory (100) operation = memory[ j ]/100; //using a switch loop to set up the loops for the cases switch ( operation ){ case 1: //reads a variable into a word from loc. Enter in -1 to exit cout <<"\n enter #: "; cin >> memory[ operand ]; break; case 2: // takes a word from location cout << "\n\nThe content at location " << operand << "is " << memory[operand]; break; case 3:// loads accum = memory[ operand ]; break; case 4: //stores memory[ operand ] = accum; break; case 5: //adds accum += mem[operand];; break; case 6: // subtracts accum-= memory[ operand ]; break; case 7: //divides accum /=(memory[ operand ]); break; case 8: // multiplies accum*= memory [ operand ]; break; case 9: // Branches to location j = operand; break; case 10: //branches if acc. is < 0 break; case 11: //branches if acc = 0 if (accum == 0) j = operand; break; case 12: // Program ends exit(0); break; } j++; } return 0; }

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