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  • Solving linear system over integers with numpy

    - by A. R. S.
    I'm trying to solve an overdetermined linear system of equations with numpy. Currently, I'm doing something like this (as a simple example): a = np.array([[1,0], [0,1], [-1,1]]) b = np.array([1,1,0]) print np.linalg.lstsq(a,b)[0] [ 1. 1.] This works, but uses floats. Is there any way to solve the system over integers only? I've tried something along the lines of print map(int, np.linalg.lstsq(a,b)[0]) [0, 1] in order to convert the solution to an array of ints, expecting [1, 1], but clearly I'm missing something. Could anyone point me in the right direction?

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  • average case running time of linear search algorithm

    - by Brahadeesh
    Hi all. I am trying to derive the average case running time for deterministic linear search algorithm. The algorithm searches an element x in an unsorted array A in the order A[1], A[2], A[3]...A[n]. It stops when it finds the element x or proceeds until it reaches the end of the array. I searched on wikipedia and the answer given was (n+1)/(k+1) where k is the number of times x is present in the array. I approached in another way and am getting a different answer. Can anyone please give me the correct proof and also let me know whats wrong with my method? E(T)= 1*P(1) + 2*P(2) + 3*P(3) ....+ n*P(n) where P(i) is the probability that the algorithm runs for 'i' time (i.e. compares 'i' elements). P(i)= (n-i)C(k-1) * (n-k)! / n! Here, (n-i)C(k-1) is (n-i) Choose (k-1). As the algorithm has reached the ith step, the rest of k-1 x's must be in the last n-i elements. Hence (n-i)C(k-i). (n-k)! is the total number of ways of arranging the rest non x numbers, and n! is the total number of ways of arranging the n elements in the array. I am not getting (n+1)/(k+1) on simplifying.

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  • linear combinations in python/numpy

    - by nmaxwell
    greetings, I'm not sure if this is a dumb question or not. Lets say I have 3 numpy arrays, A1,A2,A3, and 3 floats, c1,c2,c3 and I'd like to evaluate B = A1*c1+ A2*c2+ A3*c3 will numpy compute this as for example, E1 = A1*c1 E2 = A2*c2 E3 = A3*c3 D1 = E1+E2 B = D1+E3 or is it more clever than that? In c++ I had a neat way to abstract this kind of operation. I defined series of general 'LC' template functions, LC for linear combination like: template<class T,class D> void LC( T & R, T & L0,D C0, T & L1,D C1, T & L2,D C2) { R = L0*C0 +L1*C1 +L2*C2; } and then specialized this for various types, so for instance, for an array the code looked like for (int i=0; i<L0.length; i++) R.array[i] = L0.array[i]*C0 + L1.array[i]*C1 + L2.array[i]*C2; thus avoiding having to create new intermediate arrays. This may look messy but it worked really well. I could do something similar in python, but I'm not sure if its nescesary. Thanks in advance for any insight. -nick

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  • Triangulation & Direct linear transform

    - by srand
    Following Hartley/Zisserman's Multiview Geometery, Algorithm 12: The optimal triangulation method (p318), I got the corresponding image points xhat1 and xhat2 (step 10). In step 11, one needs to compute the 3D point Xhat. One such method is Direct Linear Transform (DLT), mentioned in 12.2 (p312) and 4.1 (p88). The homogenous method (DLT), p312-313, states that it finds a solution as the unit singular vector corresponding to the smallest singular value of A, thus, A = [xhat1(1) * P1(3,:)' - P1(1,:)' ; xhat1(2) * P1(3,:)' - P1(2,:)' ; xhat2(1) * P2(3,:)' - P2(1,:)' ; xhat2(2) * P2(3,:)' - P2(2,:)' ]; [Ua Ea Va] = svd(A); Xhat = Va(:,end); plot3(Xhat(1),Xhat(2),Xhat(3), 'r.'); However, A is a 16x1 matrix, resulting in a Va that is 1x1. What am I doing wrong (and a fix) in getting the 3D point? For what its worth sample data: xhat1 = 1.0e+009 * 4.9973 -0.2024 0.0027 xhat2 = 1.0e+011 * 2.0729 2.6624 0.0098 P1 = 699.6674 0 392.1170 0 0 701.6136 304.0275 0 0 0 1.0000 0 P2 = 1.0e+003 * -0.7845 0.0508 -0.1592 1.8619 -0.1379 0.7338 0.1649 0.6825 -0.0006 0.0001 0.0008 0.0010 A = <- my computation 1.0e+011 * -0.0000 0 0.0500 0 0 -0.0000 -0.0020 0 -1.3369 0.2563 1.5634 2.0729 -1.7170 0.3292 2.0079 2.6624

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  • Java - Using Linear Coordinates to Check Against AI [closed]

    - by Oliver Jones
    I'm working on some artificial intelligence, and I want my AI not to run into given coordinates as these are references of a wall/boundary. To begin with, every time my AI hits a wall, it makes a reference to that position (x,y). When it hits the same wall three times, it uses linear check points to 'imagine' there is a wall going through these coordinates. I want to now prevent my AI from going into that wall again. To detect if my coordinates make a straight line, i use: private boolean collinear(double x1, double y1, double x2, double y2, double x3, double y3) { return (y1 - y2) * (x1 - x3) == (y1 - y3) * (x1 - x2); } This returns true is the given points are linear to one another. So my problems are: How do I determine whether my robot is approaching the wall from its current trajectory? Instead of Java 'imagining' theres a line from 1, to 3. But to 'imagine' a line all the way through these linear coordinantes, until infinity (or close). I have a feeling this is going to require some confusing trigonometry? (REPOST: http://stackoverflow.com/questions/13542592/java-using-linear-coordinates-to-check-against-ai)

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  • Efficient algorithm to generate all solutions of a linear diophantine equation with ai=1

    - by Ben
    I am trying to generate all the solutions for the following equations for a given H. With H=4 : 1) ALL solutions for x_1 + x_2 + x_3 + x_4 =4 2) ALL solutions for x_1 + x_2 + x_3 = 4 3) ALL solutions for x_1 + x_2 = 4 4) ALL solutions for x_1 =4 For my problem, there are always 4 equations to solve (independently from the others). There are a total of 2^(H-1) solutions. For the previous one, here are the solutions : 1) 1 1 1 1 2) 1 1 2 and 1 2 1 and 2 1 1 3) 1 3 and 3 1 and 2 2 4) 4 Here is an R algorithm which solve the problem. library(gtools) H<-4 solutions<-NULL for(i in seq(H)) { res<-permutations(H-i+1,i,repeats.allowed=T) resum<-apply(res,1,sum) id<-which(resum==H) print(paste("solutions with ",i," variables",sep="")) print(res[id,]) } However, this algorithm makes more calculations than needed. I am sure it is possible to go faster. By that, I mean not generating the permutations for which the sums is H Any idea of a better algorithm for a given H ?

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  • Recommended library for linear programming in .Net?

    - by tbone
    Can anyone recommend a library - free, or commercial but affordable ( There are some listed here: http://en.wikipedia.org/wiki/Linear_programming#Solvers_and_scripting_.28programming.29_languages I am just starting out with LP and hope someone can recommend something. I am trying to basically minimize pricing for cell phone subscription services.

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  • Optimal two variable linear regression SQL statement (censoring outliers)

    - by Dave Jarvis
    Problem Am looking to apply the y = mx + b equation (where m is SLOPE, b is INTERCEPT) to a data set, which is retrieved as shown in the SQL code. The values from the (MySQL) query are: SLOPE = 0.0276653965651912 INTERCEPT = -57.2338357550468 SQL Code SELECT ((sum(t.YEAR) * sum(t.AMOUNT)) - (count(1) * sum(t.YEAR * t.AMOUNT))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as SLOPE, ((sum( t.YEAR ) * sum( t.YEAR * t.AMOUNT )) - (sum( t.AMOUNT ) * sum(power(t.YEAR, 2)))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as INTERCEPT FROM (SELECT D.AMOUNT, Y.YEAR FROM CITY C, STATION S, YEAR_REF Y, MONTH_REF M, DAILY D WHERE -- For a specific city ... -- C.ID = 8590 AND -- Find all the stations within a 15 unit radius ... -- SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) <15 AND -- Gather all known years for that station ... -- S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND -- The data before 1900 is shaky; insufficient after 2009. -- Y.YEAR BETWEEN 1900 AND 2009 AND -- Filtered by all known months ... -- M.YEAR_REF_ID = Y.ID AND -- Whittled down by category ... -- M.CATEGORY_ID = '001' AND -- Into the valid daily climate data. -- M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' GROUP BY Y.YEAR ORDER BY Y.YEAR ) t Data The data is visualized here (with five outliers highlighted): Questions How do I return the y value against all rows without repeating the same query to collect and collate the data? That is, how do I "reuse" the list of t values? How would you change the query to eliminate outliers (at an 85% confidence interval)? The following results (to calculate the start and end points of the line) appear incorrect. Why are the results off by ~10 degrees (e.g., outliers skewing the data)? (1900 * 0.0276653965651912) + (-57.2338357550468) = -4.66958228 (2009 * 0.0276653965651912) + (-57.2338357550468) = -1.65405406 I would have expected the 1900 result to be around 10 (not -4.67) and the 2009 result to be around 11.50 (not -1.65). Thank you!

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  • Optimal two variable linear regression SQL statement

    - by Dave Jarvis
    Problem Am looking to apply the y = mx + b equation (where m is SLOPE, b is INTERCEPT) to a data set, which is retrieved as shown in the SQL code. The values from the (MySQL) query are: SLOPE = 0.0276653965651912 INTERCEPT = -57.2338357550468 SQL Code SELECT ((sum(t.YEAR) * sum(t.AMOUNT)) - (count(1) * sum(t.YEAR * t.AMOUNT))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as SLOPE, ((sum( t.YEAR ) * sum( t.YEAR * t.AMOUNT )) - (sum( t.AMOUNT ) * sum(power(t.YEAR, 2)))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as INTERCEPT FROM (SELECT D.AMOUNT, Y.YEAR FROM CITY C, STATION S, YEAR_REF Y, MONTH_REF M, DAILY D WHERE -- For a specific city ... -- C.ID = 8590 AND -- Find all the stations within a 5 unit radius ... -- SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) <15 AND -- Gather all known years for that station ... -- S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND -- The data before 1900 is shaky; and insufficient after 2009. -- Y.YEAR BETWEEN 1900 AND 2009 AND -- Filtered by all known months ... -- M.YEAR_REF_ID = Y.ID AND -- Whittled down by category ... -- M.CATEGORY_ID = '001' AND -- Into the valid daily climate data. -- M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' GROUP BY Y.YEAR ORDER BY Y.YEAR ) t Data The data is visualized here: Questions How do I return the y value against all rows without repeating the same query to collect and collate the data? That is, how do I "reuse" the list of t values? How would you change the query to eliminate outliers (at an 85% confidence interval)? The following results (to calculate the start and end points of the line) appear incorrect. Why are the results off by ~10 degrees (e.g., outliers skewing the data)? (1900 * 0.0276653965651912) + (-57.2338357550468) = -4.66958228 (2009 * 0.0276653965651912) + (-57.2338357550468) = -1.65405406 I would have expected the 1900 result to be around 10 (not -4.67) and the 2009 result to be around 11.50 (not -1.65). Thank you!

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  • Optimal two variable linear regression calculation

    - by Dave Jarvis
    Problem Am looking to apply the y = mx + b equation (where m is SLOPE, b is INTERCEPT) to a data set, which is retrieved as shown in the SQL code. The values from the (MySQL) query are: SLOPE = 0.0276653965651912 INTERCEPT = -57.2338357550468 SQL Code SELECT ((sum(t.YEAR) * sum(t.AMOUNT)) - (count(1) * sum(t.YEAR * t.AMOUNT))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as SLOPE, ((sum( t.YEAR ) * sum( t.YEAR * t.AMOUNT )) - (sum( t.AMOUNT ) * sum(power(t.YEAR, 2)))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as INTERCEPT, FROM (SELECT D.AMOUNT, Y.YEAR FROM CITY C, STATION S, YEAR_REF Y, MONTH_REF M, DAILY D WHERE -- For a specific city ... -- C.ID = 8590 AND -- Find all the stations within a 15 unit radius ... -- SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) < 15 AND -- Gather all known years for that station ... -- S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND -- The data before 1900 is shaky; insufficient after 2009. -- Y.YEAR BETWEEN 1900 AND 2009 AND -- Filtered by all known months ... -- M.YEAR_REF_ID = Y.ID AND -- Whittled down by category ... -- M.CATEGORY_ID = '001' AND -- Into the valid daily climate data. -- M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' GROUP BY Y.YEAR ORDER BY Y.YEAR ) t Data The data is visualized here: Question The following results (to calculate the start and end points of the line) appear incorrect. Why are the results off by ~10 degrees (e.g., outliers skewing the data)? (1900 * 0.0276653965651912) + (-57.2338357550468) = -4.66958228 (2009 * 0.0276653965651912) + (-57.2338357550468) = -1.65405406 I would have expected the 1900 result to be around 10 (not -4.67) and the 2009 result to be around 11.50 (not -1.65). Related Sites Least absolute deviations Robust regression Thank you!

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  • Sparse linear program solver

    - by Jacob
    This great SO answer points to a good sparse solver, but I've got constraints on x (for Ax = b) such that each element in x is >=0 an <=N. The first thing which comes to mind is an LP solver for large sparse matrices. Any ideas/recommendations?

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  • A question on nature of generated assembly in C++ and code Algebra

    - by Reetesh Mukul
    I wrote this code: #include <iostream> int main() { int a; std::cin >> a; if(a*a== 3){ std::cout << a; } return 0; } On MSVC I turned ON all optimization flags. I expected that since a*a can never be 3, so compiler should not generate code for the section: if(a*a== 3){ std::cout << a; } However it generated code for the section. I did not check GCC or LLVM/CLang. What are the limits of expectation from a C++ compiler in these scenarios?

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  • Rendering Linear Gradients using the HTML5 Canvas

    - by dwahlin
    Related HTML5 Canvas Posts: Getting Started with the HTML5 Canvas Rendering Text with the HTML5 Canvas Creating a Line Chart using the HTML5 Canvas New Pluralsight Course: HTML5 Canvas Fundamentals Gradients are everywhere. They’re used to enhance toolbars or buttons and help add additional flare to a web page when used appropriately. In the past we’ve always had to rely on images to render gradients which works well, but isn’t necessarily the most efficient (although 1 pixel wide images do work well). CSS3 provides a great way to render gradients in modern browsers (see http://www.colorzilla.com/gradient-editor for a nice online gradient generator tool) but it’s not the only option. If you’re working with charts, games, multimedia or other HTML5 Canvas applications you can also use gradients and render them on the client-side without relying on images. In this post I’ll introduce how to use linear gradients and discuss the different functions that can be used to create them.   Creating Linear Gradients Linear gradients can be created using the 2D context’s createLinearGradient function. The function takes the starting x,y coordinates and ending x,y coordinates of the gradient:   createLinearGradient(x1, y1, x2, y2);   By changing the start and end coordinates you can control the direction that the gradient renders. For example, adding the following coordinates causes the gradient to render from left to right since the y value stays at 0 for both points while the x value changes from 0 to 200. var lgrad = ctx.createLinearGradient(0, 0, 200, 0); Here’s an example of how changing the coordinates affects the gradient direction:   Once a linear gradient object has been created you can set color stops using the addColorStop() function. It takes the location where the color should appear in the gradient with 0 being the beginning and 1 being at the end (0.5 would be in the middle) as well as the color to display in the gradient. lgrad.addColorStop(0, 'white'); lgrad.addColorStop(1, 'gray');   An example of combining createLinearGradient() with addColorStop() is shown next:   Using createLinearGradient() var canvas = document.getElementById('myCanvas'); var ctx = canvas.getContext('2d'); var lgrad = ctx.createLinearGradient(0, 0, 200, 0); lgrad.addColorStop(0, 'white'); lgrad.addColorStop(1, 'gray'); ctx.fillStyle = lgrad; ctx.fillRect(0, 0, 200, 200); ctx.strokeRect(0, 0, 200, 200); This code renders a white to gray gradient as shown next: A live example of using createLinearGradient() is shown next. Click the Result tab to see the code in action.   In the next post on the HTML5 Canvas I’ll take a look at radial gradients and how they can be used. In the meantime, if you’re interested in learning more about the HTML5 Canvas and how it can be used in your Web or Windows 8 applications, check out my HTML5 Canvas Fundamentals course from Pluralsight. It has over 4 1/2 hours of canvas goodness packed in it.

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  • How to set position for a linear-gradient background in css3

    - by Virender Sehwag
    I am trying to set the position (that is, margin or padding from top) of body tag's linear background with image. My code is background-image: linear-gradient(to bottom, rgba(255, 255, 255, 0), rgba(255, 255, 255, 0), rgba(0, 0, 0, 0.9), rgb(0, 0, 0)), url("g2hd.jpg"); background-repeat: repeat, no-repeat; background-attachment: fixed; background-position: 0% 30px, center center; but 30px is not working but it works for normal for background-image:url("g2hd.jpg"); any idea

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  • How to solve generic algebra using solver/library programmatically? Matlab, Mathematica, Wolfram etc?

    - by DevDevDev
    I'm trying to build an algebra trainer for students. I want to construct a representative problem, define constraints and relationships on the parameters, and then generate a bunch of Latex formatted problems from the representation. As an example: A specific question might be: If y < 0 and (x+3)(y-5) = 0, what is x? Answer (x = -3) I would like to encode this as a Latex formatted problem like. If $y<0$ and $(x+constant_1)(y+constant_2)=0$ what is the value of x? Answer = -constant_1 And plug into my problem solver constant_1 > 0, constant_1 < 60, constant_1 = INTEGER constant_2 < 0, constant_2 > -60, constant_2 = INTEGER Then it will randomly construct me pairs of (constant_1, constant_2) that I can feed into my Latex generator. Obviously this is an extremely simple example with no real "solving" but hopefully it gets the point across. Things I'm looking for ideally in priority order * Solve algebra problems * Definition of relationships relatively straight forward * Rich support for latex formatting (not just writing encoded strings) Thanks!

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  • linear algebra libraries for clusters

    - by Abruzzo Forte e Gentile
    Hi all I need to develop applications doing linear algebra + eigenvalue + linear equation solutions over a cluster of pcs ( I have a lot of machines available ). I discovered Scalapack libraries but they seem to me developed long time ago. Do you know if these are other libs available that I should learn doing math & linear algebra in a cluster? My language is C++ and off course I am newbie to this topic. Kind Regards to everybody AFG

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  • How do you calculate the reflex angle given to vectors in 3D space?

    - by Reimund
    I want to calculate the angle between two vectors a and b. Lets assume these are at the origin. This can be done with theta = arccos(a . b / |a| * |b|) However arccos gives you the angle in [0, pi], i.e. it will never give you an angle greater than 180 degrees, which is what I want. So how do you find out when the vectors have gone past the 180 degree mark? In 2D I would simply let the sign of the y-component on one of the vectors determine what quadrant the vector is in. But what is the easiest way to do it in 3D?

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  • Sparse quadratic program solver

    - by Jacob
    This great SO answer points to a good sparse solver, but I've got constraints on x (for Ax = b) such that each element in x is >=0 an <=N. The first thing which comes to mind is an QP solver for large sparse matrices. Also, A is huge (around 2e6x2e6) but very sparse with <=4 elements per row. Any ideas/recommendations?

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  • Create a linear trend line in Excel graphs with logarithmic scale

    - by Redsoft7
    I I have an Excel scatter chart with x and y values. I set the logarithmic scale in x-axis and y-axis. When I add a linear trend line to the graph, the line is not linear but appears like a curve. How can I make a linear trend line on a logarithmic-scaled chart? Sample data: x: 18449 22829 25395 36869 101419 125498 208144 2001508 14359478 17301785 y: 269,09 273,89 239,50 239,50 175,13 176,73 151,94 135,15 131,55 121,55

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  • Increasing speed of circle over time as linear with Box2d

    - by Whispered
    Assume that there is a circle and it can be moved by using keyboard arrows.Is required that increasing speed over time like increasing car speed. For example; max speed is 25 and time to reach max speed shall be 5 sec. Over 5 sec the speed will reach to max speed. Does Box2d handle that situation?. I tried setting linear valocity but it seems to make the circle have constant speed instead of increased speed over time. Thank You! Note: I'm using Box2DWeb Javascript port of Box2D.

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  • How to solve linear recurrences involving two functions?

    - by Aditya Bahuguna
    Actually I came across a question in Dynamic Programming where we need to find the number of ways to tile a 2 X N area with tiles of given dimensions.. Here is the problem statement Now after a bit of recurrence solving I came out with these. F(n) = F(n-1) + F(n-2) + 2G(n-1), and G(n) = G(n-1) + F(n-1) I know how to solve LR model where one function is there.For large N as is the case in the above problem we can do the matrix exponentiation and achieve O(k^3log(N)) time where k is the minimum number such that for all km F(n) does not depend on F(n-k). The method of solving linear recurrence with matrix exponentiation as it is given in that blog. Now for the LR involving two functions can anyone suggest an approach feasible enough for large N.

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  • Show path of a body of where it should go after linear impulse is applied

    - by Farooq Arshed
    I am making a game with Andengine and Box2D. I have a dynamic body and I apply linear impulse on the body to move it around when the user have touched the screen. Now I want to show the path where the body will go when the user have touched. If you have played Angry Birds or Basket Ball Shoot or any other which have projectile motion with a path shown you will get my point. I want to show the white dots which are shown in those games.

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  • -webkit-linear-gradient Not working in Dreamweaver CS6

    - by Ken
    I've tried multiple times to apply the following piece of code in a CSS document: display: block; width: 500px; margin: 500px auto; padding: 15px; text-align: center; border: 4px solid blue; background: -webkit-linear-gradient(top,black,white); outline: 7px solid red; Everything appears correctly, except the gradient. I have CS6 Live View turned on, and it still refuses to appear. All I get in my box is a white background, as opposed to the black to white gradient. However, when I type the same line of code into the trial of Coda 2 I downloaded, it works perfectly. Is there anything I can do to resolve the issue?

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