Search Results

Search found 1430 results on 58 pages for 'moz linear gradient'.

Page 3/58 | < Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >

  • Haskell Linear Algebra Matrix Library for Arbitrary Element Types

    - by Johannes Weiß
    I'm looking for a Haskell linear algebra library that has the following features: Matrix multiplication Matrix addition Matrix transposition Rank calculation Matrix inversion is a plus and has the following properties: arbitrary element (scalar) types (in particular element types that are not Storable instances). My elements are an instance of Num, additionally the multiplicative inverse can be calculated. The elements mathematically form a finite field (??2256). That should be enough to implement the features mentioned above. arbitrary matrix sizes (I'll probably need something like 100x100, but the matrix sizes will depend on the user's input so it should not be limited by anything else but the memory or the computational power available) as fast as possible, but I'm aware that a library for arbitrary elements will probably not perform like a C/Fortran library that does the work (interfaced via FFI) because of the indirection of arbitrary (non Int, Double or similar) types. At least one pointer gets dereferenced when an element is touched (written in Haskell, this is not a real requirement for me, but since my elements are no Storable instances the library has to be written in Haskell) I already tried very hard and evaluated everything that looked promising (most of the libraries on Hackage directly state that they wont work for me). In particular I wrote test code using: hmatrix, assumes Storable elements Vec, but the documentation states: Low Dimension : Although the dimensionality is limited only by what GHC will handle, the library is meant for 2,3 and 4 dimensions. For general linear algebra, check out the excellent hmatrix library and blas bindings I looked into the code and the documentation of many more libraries but nothing seems to suit my needs :-(. Update Since there seems to be nothing, I started a project on GitHub which aims to develop such a library. The current state is very minimalistic, not optimized for speed at all and only the most basic functions have tests and therefore should work. But should you be interested in using or helping out developing it: Contact me (you'll find my mail address on my web site) or send pull requests.

    Read the article

  • how to Compute the average probe length for success and failure - Linear probe (Hash Tables)

    - by fang_dejavu
    hi everyone, I'm doing an assignment for my Data Structures class. we were asked to to study linear probing with load factors of .1, .2 , .3, ...., and .9. The formula for testing is: The average probe length using linear probing is roughly Success-- ( 1 + 1/(1-L)**2)/2 or Failure-- (1+1(1-L))/2. we are required to find the theoretical using the formula above which I did(just plug the load factor in the formula), then we have to calculate the empirical (which I not quite sure how to do). here is the rest of the requirements **For each load factor, 10,000 randomly generated positive ints between 1 and 50000 (inclusive) will be inserted into a table of the "right" size, where "right" is strictly based upon the load factor you are testing. Repeats are allowed. Be sure that your formula for randomly generated ints is correct. There is a class called Random in java.util. USE it! After a table of the right (based upon L) size is loaded with 10,000 ints, do 100 searches of newly generated random ints from the range of 1 to 50000. Compute the average probe length for each of the two formulas and indicate the denominators used in each calculationSo, for example, each test for a .5 load would have a table of size approximately 20,000 (adjusted to be prime) and similarly each test for a .9 load would have a table of approximate size 10,000/.9 (again adjusted to be prime). The program should run displaying the various load factors tested, the average probe for each search (the two denominators used to compute the averages will add to 100), and the theoretical answers using the formula above. .** how do I calculate the empirical success?

    Read the article

  • how does linear probing handle this?

    - by Weadadada Awda
    • the hash function: h(x) = | 2x + 5 | mod M • a bucket array of capacity N • a set of objects with keys: 12, 44, 13, 88, 23, 94, 11, 39, 20, 16, 5 (to input from left to right) 4.a [5 pts] Write the hash table where M=N=11 and collisions are handled using linear probing. So I got up to here x x x x x 44 88 12 23 13 94 but the next variable should go after the 94 now, (the 11) but does it start from the beggining or what? thx

    Read the article

  • Linear Regression and Java Dates

    - by Smithers
    I am trying to find the linear trend line for a set of data. The set contains pairs of dates (x values) and scores (y values). I am using a version of this code as the basis of my algorithm. The results I am getting are off by a few orders of magnitude. I assume that there is some problem with round off error or overflow because I am using Date's getTime method which gives you a huge number of milliseconds. Does anyone have a suggestion on how to minimize the errors and compute the correct results?

    Read the article

  • 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?

    Read the article

  • Quand Chrome gagne 40 millions d'utilisateurs, Firefox en gagne 100 millions d'après un cadre de Moz

    Mise à jour du 21/05/10 Quand Chrome gagne 40 millions d'utilisateurs, Firefox en gagne 100 D'après un cadre de Mozilla : qui parle de déclin ? Comme d'habitude avec la Fondation Mozilla, il ne s'agit pas d'une réponse officielle. Mais cela y ressemble furieusement. Sur son blog personnel, Asa Dotzler, directeur du développement de Firefox, vient de comparer les progressions respectives de Chrome et de Firefox sur l'année 2009. Cette mini-étude fait suite aux déclarations de Black Ross, un des créateurs du navigateur, pour qui le Panda Roux est proche du déclin et la Fondation empêtrée dans une culture bureaucratique qu...

    Read the article

  • 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.

    Read the article

  • 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

    Read the article

  • 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

    Read the article

  • How do you change color of gradient image found on the net?

    - by askmoo
    For example, I found this gradient image (randomly chosen) How do you change its color in Photoshop or GIMP? I tried overlay but it covers everything with that color. For example I want to have white-to-red gradient image. Possible to do it in a quick way via any of these tools or I have to make it from the scratch? This is the image before and after I use the tool suggested. You can see the horizontal striped on the after image.

    Read the article

  • CSS3 gradient background set on body doesn't stretch but instead repeats?

    - by John Isaacks
    ok say the content inside the <body> totals 300px high. If I set the background of my <body> using -webkit-gradient or -moz-linear-gradient Then I maximize my window (or just make it taller than 300px) the gradient will be exactly 300px tall (the height of the content) and just repeat to fill the rest of the window. I am assuming this is not a bug since it is the same in both webkit and gecko. But is there a way to make the gradient stretch to fill the window instead of repeat?

    Read the article

  • Gradient and window re sizing with css [migrated]

    - by guisasso
    The situation: A table with width set to 100%, that has a cell inside with 1000px width. The table is centered, and so is the cell. I would like to have a gradient from left to right, and right to left that would end at the beginning of the centered cell, with the same color as the cell. The problem is, to occupy the whole page, no matter what size the browser is, the table is set to 100%, the cell is set to 1000px so it'll never change its size, How can i achieve, if possible, what i want, making sure that in smaller resolutions/monitors or with window re sizing, the gradient will stop at the beginning of that cell, since gradients are set with percentage?

    Read the article

  • 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)

    Read the article

  • 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

    Read the article

  • 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 ?

    Read the article

  • 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.

    Read the article

  • 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!

    Read the article

  • 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!

    Read the article

  • 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!

    Read the article

  • 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?

    Read the article

  • O'Reilly book clarification on 2d linear system

    - by Eric
    The Oreilly book "Learning openCV" states at page 356 : Quote Before we get totally lost, let’s consider a particular realistic situation of taking measurements on a car driving in a parking lot. We might imagine that the state of the car could be summarized by two position variables, x and y, and two velocities, vx and vy. These four variables would be the elements of the state vector xk. Th is suggests that the correct form for F is: x = [ x; y; vx; vy; ]k F = [ 1, 0, dt, 0; 0, 1, 0, dt; 0, 0, 1, 0; 0, 0, 0, 1; ] It seems natural to put 'dt' just there in the F matrix but I just don't get why. What if I have a n states system, how would I spray some "dt" in the F matrix?

    Read the article

  • -webkit- vs -moz-transition

    - by danixd
    I am using CSS3 transitions on my site and the -webkit- seems to be working, whilst the -moz- is not. Here is the CSS: article {z-index: 2; float: left; overflow: hidden; position: relative; -webkit-transition: -webkit-transform 0.2s ease-in-out; -moz-transition: -moz-transform 0.2s ease-in-out; } .mousedown{-webkit-transform: translate(-180px, 0) !important; -moz-transform: translate(-180px, 0) !important; } Just using jQeury to add the mousedown class onto the article. Any idea where I am going wrong?

    Read the article

< Previous Page | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >