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  • Scheduling thread tiles with C++ AMP

    - by Daniel Moth
    This post assumes you are totally comfortable with, what some of us call, the simple model of C++ AMP, i.e. you could write your own matrix multiplication. We are now ready to explore the tiled model, which builds on top of the non-tiled one. Tiling the extent We know that when we pass a grid (which is just an extent under the covers) to the parallel_for_each call, it determines the number of threads to schedule and their index values (including dimensionality). For the single-, two-, and three- dimensional cases you can go a step further and subdivide the threads into what we call tiles of threads (others may call them thread groups). So here is a single-dimensional example: extent<1> e(20); // 20 units in a single dimension with indices from 0-19 grid<1> g(e);      // same as extent tiled_grid<4> tg = g.tile<4>(); …on the 3rd line we subdivided the single-dimensional space into 5 single-dimensional tiles each having 4 elements, and we captured that result in a concurrency::tiled_grid (a new class in amp.h). Let's move on swiftly to another example, in pictures, this time 2-dimensional: So we start on the left with a grid of a 2-dimensional extent which has 8*6=48 threads. We then have two different examples of tiling. In the first case, in the middle, we subdivide the 48 threads into tiles where each has 4*3=12 threads, hence we have 2*2=4 tiles. In the second example, on the right, we subdivide the original input into tiles where each has 2*2=4 threads, hence we have 4*3=12 tiles. Notice how you can play with the tile size and achieve different number of tiles. The numbers you pick must be such that the original total number of threads (in our example 48), remains the same, and every tile must have the same size. Of course, you still have no clue why you would do that, but stick with me. First, we should see how we can use this tiled_grid, since the parallel_for_each function that we know expects a grid. Tiled parallel_for_each and tiled_index It turns out that we have additional overloads of parallel_for_each that accept a tiled_grid instead of a grid. However, those overloads, also expect that the lambda you pass in accepts a concurrency::tiled_index (new in amp.h), not an index<N>. So how is a tiled_index different to an index? A tiled_index object, can have only 1 or 2 or 3 dimensions (matching exactly the tiled_grid), and consists of 4 index objects that are accessible via properties: global, local, tile_origin, and tile. The global index is the same as the index we know and love: the global thread ID. The local index is the local thread ID within the tile. The tile_origin index returns the global index of the thread that is at position 0,0 of this tile, and the tile index is the position of the tile in relation to the overall grid. Confused? Here is an example accompanied by a picture that hopefully clarifies things: array_view<int, 2> data(8, 6, p_my_data); parallel_for_each(data.grid.tile<2,2>(), [=] (tiled_index<2,2> t_idx) restrict(direct3d) { /* todo */ }); Given the code above and the picture on the right, what are the values of each of the 4 index objects that the t_idx variables exposes, when the lambda is executed by T (highlighted in the picture on the right)? If you can't work it out yourselves, the solution follows: t_idx.global       = index<2> (6,3) t_idx.local          = index<2> (0,1) t_idx.tile_origin = index<2> (6,2) t_idx.tile             = index<2> (3,1) Don't move on until you are comfortable with this… the picture really helps, so use it. Tiled Matrix Multiplication Example – part 1 Let's paste here the C++ AMP matrix multiplication example, bolding the lines we are going to change (can you guess what the changes will be?) 01: void MatrixMultiplyTiled_Part1(vector<float>& vC, const vector<float>& vA, const vector<float>& vB, int M, int N, int W) 02: { 03: 04: array_view<const float,2> a(M, W, vA); 05: array_view<const float,2> b(W, N, vB); 06: array_view<writeonly<float>,2> c(M, N, vC); 07: parallel_for_each(c.grid, 08: [=](index<2> idx) restrict(direct3d) { 09: 10: int row = idx[0]; int col = idx[1]; 11: float sum = 0.0f; 12: for(int i = 0; i < W; i++) 13: sum += a(row, i) * b(i, col); 14: c[idx] = sum; 15: }); 16: } To turn this into a tiled example, first we need to decide our tile size. Let's say we want each tile to be 16*16 (which assumes that we'll have at least 256 threads to process, and that c.grid.extent.size() is divisible by 256, and moreover that c.grid.extent[0] and c.grid.extent[1] are divisible by 16). So we insert at line 03 the tile size (which must be a compile time constant). 03: static const int TS = 16; ...then we need to tile the grid to have tiles where each one has 16*16 threads, so we change line 07 to be as follows 07: parallel_for_each(c.grid.tile<TS,TS>(), ...that means that our index now has to be a tiled_index with the same characteristics as the tiled_grid, so we change line 08 08: [=](tiled_index<TS, TS> t_idx) restrict(direct3d) { ...which means, without changing our core algorithm, we need to be using the global index that the tiled_index gives us access to, so we insert line 09 as follows 09: index<2> idx = t_idx.global; ...and now this code just works and it is tiled! Closing thoughts on part 1 The process we followed just shows the mechanical transformation that can take place from the simple model to the tiled model (think of this as step 1). In fact, when we wrote the matrix multiplication example originally, the compiler was doing this mechanical transformation under the covers for us (and it has additional smarts to deal with the cases where the total number of threads scheduled cannot be divisible by the tile size). The point is that the thread scheduling is always tiled, even when you use the non-tiled model. But with this mechanical transformation, we haven't gained anything… Hint: our goal with explicitly using the tiled model is to gain even more performance. In the next post, we'll evolve this further (beyond what the compiler can automatically do for us, in this first release), so you can see the full usage of the tiled model and its benefits… Comments about this post by Daniel Moth welcome at the original blog.

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  • How to calculate the covariance in T-SQL

    - by Peter Larsson
    DECLARE @Sample TABLE         (             x INT NOT NULL,             y INT NOT NULL         ) INSERT  @Sample VALUES  (3, 9),         (2, 7),         (4, 12),         (5, 15),         (6, 17) ;WITH cteSource(x, xAvg, y, yAvg, n) AS (         SELECT  1E * x,                 AVG(1E * x) OVER (PARTITION BY (SELECT NULL)),                 1E * y,                 AVG(1E * y) OVER (PARTITION BY (SELECT NULL)),                 COUNT(*) OVER (PARTITION BY (SELECT NULL))         FROM    @Sample ) SELECT  SUM((x - xAvg) *(y - yAvg)) / MAX(n) AS [COVAR(x,y)] FROM    cteSource

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  • How to discriminate from two nodes with identical frequencies in a Huffman's tree?

    - by Omega
    Still on my quest to compress/decompress files with a Java implementation of Huffman's coding (http://en.wikipedia.org/wiki/Huffman_coding) for a school assignment. From the Wikipedia page, I quote: Create a leaf node for each symbol and add it to the priority queue. While there is more than one node in the queue: Remove the two nodes of highest priority (lowest probability) from the queue Create a new internal node with these two nodes as children and with probability equal to the sum of the two nodes' probabilities. Add the new node to the queue. The remaining node is the root node and the tree is complete. Now, emphasis: Remove the two nodes of highest priority (lowest probability) from the queue Create a new internal node with these two nodes as children and with probability equal to the sum of the two nodes' probabilities. So I have to take two nodes with the lowest frequency. What if there are multiple nodes with the same low frequency? How do I discriminate which one to use? The reason I ask this is because Wikipedia has this image: And I wanted to see if my Huffman's tree was the same. I created a file with the following content: aaaaeeee nnttmmiihhssfffouxprl And this was the result: Doesn't look so bad. But there clearly are some differences when multiple nodes have the same frequency. My questions are the following: What is Wikipedia's image doing to discriminate the nodes with the same frequency? Is my tree wrong? (Is Wikipedia's image method the one and only answer?) I guess there is one specific and strict way to do this, because for our school assignment, files that have been compressed by my program should be able to be decompressed by other classmate's programs - so there must be a "standard" or "unique" way to do it. But I'm a bit lost with that. My code is rather straightforward. It literally just follows Wikipedia's listed steps. The way my code extracts the two nodes with the lowest frequency from the queue is to iterate all nodes and if the current node has a lower frequency than any of the two "smallest" known nodes so far, then it replaces the highest one. Just like that.

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  • How to remove the boundary effects arising due to zero padding in scipy/numpy fft?

    - by Omkar
    I have made a python code to smoothen a given signal using the Weierstrass transform, which is basically the convolution of a normalised gaussian with a signal. The code is as follows: #Importing relevant libraries from __future__ import division from scipy.signal import fftconvolve import numpy as np def smooth_func(sig, x, t= 0.002): N = len(x) x1 = x[-1] x0 = x[0] # defining a new array y which is symmetric around zero, to make the gaussian symmetric. y = np.linspace(-(x1-x0)/2, (x1-x0)/2, N) #gaussian centered around zero. gaus = np.exp(-y**(2)/t) #using fftconvolve to speed up the convolution; gaus.sum() is the normalization constant. return fftconvolve(sig, gaus/gaus.sum(), mode='same') If I run this code for say a step function, it smoothens the corner, but at the boundary it interprets another corner and smoothens that too, as a result giving unnecessary behaviour at the boundary. I explain this with a figure shown in the link below. Boundary effects This problem does not arise if we directly integrate to find convolution. Hence the problem is not in Weierstrass transform, and hence the problem is in the fftconvolve function of scipy. To understand why this problem arises we first need to understand the working of fftconvolve in scipy. The fftconvolve function basically uses the convolution theorem to speed up the computation. In short it says: convolution(int1,int2)=ifft(fft(int1)*fft(int2)) If we directly apply this theorem we dont get the desired result. To get the desired result we need to take the fft on a array double the size of max(int1,int2). But this leads to the undesired boundary effects. This is because in the fft code, if size(int) is greater than the size(over which to take fft) it zero pads the input and then takes the fft. This zero padding is exactly what is responsible for the undesired boundary effects. Can you suggest a way to remove this boundary effects? I have tried to remove it by a simple trick. After smoothening the function I am compairing the value of the smoothened signal with the original signal near the boundaries and if they dont match I replace the value of the smoothened func with the input signal at that point. It is as follows: i = 0 eps=1e-3 while abs(smooth[i]-sig[i])> eps: #compairing the signals on the left boundary smooth[i] = sig[i] i = i + 1 j = -1 while abs(smooth[j]-sig[j])> eps: # compairing on the right boundary. smooth[j] = sig[j] j = j - 1 There is a problem with this method, because of using an epsilon there are small jumps in the smoothened function, as shown below: jumps in the smooth func Can there be any changes made in the above method to solve this boundary problem?

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  • is there a formal algebra method to analyze programs?

    - by Gabriel
    Is there a formal/academic connection between an imperative program and algebra, and if so where would I learn about it? The example I'm thinking of is: if(C1) { A1(); A2(); } if(C2) { A1(); A2(); } Represented as a sum of terms: (C1)(A1) + (C1)(A2) + (C2)(A1) + (C2)(A2) = (C1+C2)(A1+A2) The idea being that manipulation could lead to programatic refactoring - "factoring" being the common concept in this example.

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  • how to fix it =.= "Third party sources disabled"

    - by Long Pham
    When i tried to upgrade to 13.10, it was appearing on the screen while preparing to update "Third party sources disabled Some third party entries in your sources.list were disabled. You can re-enable them after the upgrade with the 'software-properties' tool or your package manager." After that it was on: W:Failed to fetch bzip2:/var/lib/apt/lists/partial/archive.ubuntu.com_ubuntu_dists_saucy_universe_source_Sources Hash Sum mismatch , E:Some index files failed to download. They have been ignored, or old ones used instead.

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  • algorithm for project euler problem no 18

    - by Valentino Ru
    Problem number 18 from Project Euler's site is as follows: By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23. 3 7 4 2 4 6 8 5 9 3 That is, 3 + 7 + 4 + 9 = 23. Find the maximum total from top to bottom of the triangle below: 75 95 64 17 47 82 18 35 87 10 20 04 82 47 65 19 01 23 75 03 34 88 02 77 73 07 63 67 99 65 04 28 06 16 70 92 41 41 26 56 83 40 80 70 33 41 48 72 33 47 32 37 16 94 29 53 71 44 65 25 43 91 52 97 51 14 70 11 33 28 77 73 17 78 39 68 17 57 91 71 52 38 17 14 91 43 58 50 27 29 48 63 66 04 68 89 53 67 30 73 16 69 87 40 31 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23 NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o) The formulation of this problems does not make clear if the "Traversor" is greedy, meaning that he always choosed the child with be higher value the maximum of every single walkthrough is asked The NOTE says, that it is possible to solve this problem by trying every route. This means to me, that is is also possible without! This leads to my actual question: Assumed that not the greedy one is the max, is there any algorithm that finds the max walkthrough value without trying every route and that doesn't act like the greedy algorithm? I implemented an algorithm in Java, putting the values first in a node structure, then applying the greedy algorithm. The result, however, is cosidered as wrong by Project Euler. sum = 0; void findWay(Node node){ sum += node.value; if(node.nodeLeft != null && node.nodeRight != null){ if(node.nodeLeft.value > node.nodeRight.value){ findWay(node.nodeLeft); }else{ findWay(node.nodeRight); } } }

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  • Parent-child hierarchies and unary operators in PowerPivot

    - by Marco Russo (SQLBI)
    Alberto wrote an excellent post describing how to implement the Unary Operator feature (which is present in Analysis Services) in PowerPivot (there was a previous post about parent-child hierarchies, too). I have to say that the solution is not so easy to implement as in Analysis Services, but it just works and, from a practical point of view, it is not so difficult to implement if you understand how it works and accept its limitations (only sum and subtractions are supported). I think that many...(read more)

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  • How would you go about tackling this problem? [SOLVED in C++]

    - by incrediman
    Intro: EDIT: See solution at the bottom of this question (c++) I have a programming contest coming up in about half a week, and I've been prepping :) I found a bunch of questions from this canadian competition, they're great practice: http://cemc.math.uwaterloo.ca/contests/computing/2009/stage2/day1.pdf I'm looking at problem B ("Dinner"). Any idea where to start? I can't really think of anything besides the naive approach (ie. trying all permutations) which would take too long to be a valid answer. Btw, the language there says c++ and pascal I think, but i don't care what language you use - I mean really all I want is a hint as to the direction I should proceed in, and perhpas a short explanation to go along with it. It feels like I'm missing something obvious... Of course extended speculation is more than welcome, but I just wanted to clarify that I'm not looking for a full solution here :) Short version of the question: You have a binary string N of length 1-100 (in the question they use H's and G's instead of one's and 0's). You must remove all of the digits from it, in the least number of steps possible. In each step you may remove any number of adjacent digits so long as they are the same. That is, in each step you can remove any number of adjacent G's, or any number of adjacent H's, but you can't remove H's and G's in one step. Example: HHHGHHGHH Solution to the example: 1. HHGGHH (remove middle Hs) 2. HHHH (remove middle Gs) 3. Done (remove Hs) -->Would return '3' as the answer. Note that there can also be a limit placed on how large adjacent groups have to be when you remove them. For example it might say '2', and then you can't remove single digits (you'd have to remove pairs or larger groups at a time). Solution I took Mark Harrison's main algorithm, and Paradigm's grouping idea and used them to create the solution below. You can try it out on the official test cases if you want. //B.cpp //include debug messages? #define DEBUG false #include <iostream> #include <stdio.h> #include <vector> using namespace std; #define FOR(i,n) for (int i=0;i<n;i++) #define FROM(i,s,n) for (int i=s;i<n;i++) #define H 'H' #define G 'G' class String{ public: int num; char type; String(){ type=H; num=0; } String(char type){ this->type=type; num=1; } }; //n is the number of bits originally in the line //k is the minimum number of people you can remove at a time //moves is the counter used to determine how many moves we've made so far int n, k, moves; int main(){ /*Input from File*/ scanf("%d %d",&n,&k); char * buffer = new char[200]; scanf("%s",buffer); /*Process input into a vector*/ //the 'line' is a vector of 'String's (essentially contigious groups of identical 'bits') vector<String> line; line.push_back(String()); FOR(i,n){ //if the last String is of the correct type, simply increment its count if (line.back().type==buffer[i]) line.back().num++; //if the last String is of the wrong type but has a 0 count, correct its type and set its count to 1 else if (line.back().num==0){ line.back().type=buffer[i]; line.back().num=1; } //otherwise this is the beginning of a new group, so create the new group at the back with the correct type, and a count of 1 else{ line.push_back(String(buffer[i])); } } /*Geedily remove groups until there are at most two groups left*/ moves=0; int I;//the position of the best group to remove int bestNum;//the size of the newly connected group the removal of group I will create while (line.size()>2){ /*START DEBUG*/ if (DEBUG){ cout<<"\n"<<moves<<"\n----\n"; FOR(i,line.size()) printf("%d %c \n",line[i].num,line[i].type); cout<<"----\n"; } /*END DEBUG*/ I=1; bestNum=-1; FROM(i,1,line.size()-1){ if (line[i-1].num+line[i+1].num>bestNum && line[i].num>=k){ bestNum=line[i-1].num+line[i+1].num; I=i; } } //remove the chosen group, thus merging the two adjacent groups line[I-1].num+=line[I+1].num; line.erase(line.begin()+I);line.erase(line.begin()+I); moves++; } /*START DEBUG*/ if (DEBUG){ cout<<"\n"<<moves<<"\n----\n"; FOR(i,line.size()) printf("%d %c \n",line[i].num,line[i].type); cout<<"----\n"; cout<<"\n\nFinal Answer: "; } /*END DEBUG*/ /*Attempt the removal of the last two groups, and output the final result*/ if (line.size()==2 && line[0].num>=k && line[1].num>=k) cout<<moves+2;//success else if (line.size()==1 && line[0].num>=k) cout<<moves+1;//success else cout<<-1;//not everyone could dine. /*START DEBUG*/ if (DEBUG){ cout<<" moves."; } /*END DEBUG*/ }

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  • SAF Architecture Evaluation (Introduction)

    I saidin “what’s Software architecture” – architecture is both an early artifact and it also represents the significant decisions about the system – or to sum it up”Architecture is the decisions that youwishyou could get right early in a project.(Ralph Johnston*). That is exactly why I made evaluation one of the key steps in SAF. [...]...Did you know that DotNetSlackers also publishes .net articles written by top known .net Authors? We already have over 80 articles in several categories including Silverlight. Take a look: here.

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  • Oracle Snapshot Not Working [closed]

    - by nayef harb
    i have created a snapshot that takes data from 2 tables and has a refresh rate of 1 day. The snapshot data is not refreshing it is still the same. is there something that i am missing ? Here is the code: CREATE SNAPSHOT test REFRESH COMPLETE START WITH SYSDATE NEXT sysdate + 1 AS select item_code,item_conc_code,tran_bran_code,sum(tran_qty) bal_qty from tranhist a, itemmast b where a.tran_item_code = b.item_code group by item_code,item_conc_code,tran_bran_code

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  • Better way to design a database

    - by cMinor
    I have a conceptual problem and I would like to get your ideas on how I'll be able to do what I am aiming. My goal is to create a database with information of persons who work at a place depending on their profession and skills,and keep control of salary and projects (how much would cost summing all the hours of work) I have 3 categories which can have subcategories: Outsourcing Technician welder turner assistant Administrative supervisor manager So each person has its information and the projects they are working on, also one person may do several jobs... I was thinking about having 5 tables (EMPLOYEE, SKILLS, PROYECTS, SALARY, PROFESSION) but I guess there is a better way of doing this. create table Employee ( PRIMARY KEY [Person_ID] int(10), [Name] varchar(30), [sex] varchar(10), [address] varchar(10), [profession] varchar(10), [Skills_ID] int(10), [Proyect_ID] int(10), [Salary_ID] int(10), [Salary] float ) create table Skills ( PRIMARY KEY [Skills_ID] int(10), FOREIGN KEY [Skills_name] varchar(10) REFERENCES Employee(Person_ID), [Skills_pay] float(10), [Comments] varchar(50) ) create table Proyects ( PRIMARY KEY [Proyect_ID] int(10), FOREIGN KEY [Skills_name] varchar(10) REFERENCES Employee(Person_ID) [Proyect_name] varchar(10), [working_Hours] float(10), [Comments] varchar(50) ) create table Salary ( PRIMARY KEY [Salary_ID] int(10), FOREIGN KEY [Skills_name] varchar(10) REFERENCES Employee(Person_ID) [Proyect_name] varchar(10), [working_Hours] float(10), [Comments] varchar(50) ) So to get the total amount of the cost of a project I would just sum the working hours of each employee envolved and sum some extra costs in an aggregate query. Is there a way to do this in a more efficient way? What to add or delete of this small model? I guess I am missing something in the salary - maybe I need another table for that?

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  • Excel not properly recalculating values

    - by gms8994
    I have an excel sheet with values in it (this sheet is generated by a custom perl script, but I don't think that's where the problem lies). In it, I have a formula: =sum(indirect(concatenate(address(6,column()),":",address(17,column())))) The purpose of this formula is to give me the SUM() of the cells in the current column, between rows 6 and 17. In Gnumeric Spreadsheet, as soon as I open the file, this works. But in Excel (both 2003 and 2007), opening the file gives #VALUE! errors in the fields with this formula, stating that the INDIRECT call with the values $B$6:$B$17 will result in an error. Here's the kink in the issue. If I edit the field (via F2), and make no changes, and hit enter, the values update. Also, it seems, if I save the file as .xlsx (Excel 2007 format), the values update upon opening. Unfortunately, I'm not sure that creating an xlsx is a possibility with the modules that I'm using, and many of our clients probably wouldn't be able to use it anyway. Any suggestions? Editing 200+ files every month for each client isn't going to be feasible, so if there's something I'm missing, please let me know.

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  • MySQL select query result set changes based on column order

    - by user197191
    I have a drupal 7 site using the Views module to back-end site content search results. The same query with the same dataset returns different results from MySQL 5.5.28 to MySQL 5.6.14. The results from 5.5.28 are the correct, expected results. The results from 5.6.14 are not. If, however, I simply move a column in the select statement, the query returns the correct results. Here is the code-generated query in question (modified for readability). I apologize for the length; I couldn't find a way to reproduce it without the whole query: SELECT DISTINCT node_node_revision.nid AS node_node_revision_nid, node_revision.title AS node_revision_title, node_field_revision_field_position_institution_ref.nid AS node_field_revision_field_position_institution_ref_nid, node_revision.vid AS vid, node_revision.nid AS node_revision_nid, node_node_revision.title AS node_node_revision_title, SUM(search_index.score * search_total.count) AS score, 'node' AS field_data_field_system_inst_name_node_entity_type, 'node' AS field_revision_field_position_college_division_node_entity_t, 'node' AS field_revision_field_position_department_node_entity_type, 'node' AS field_revision_field_search_lvl_degree_lvls_node_entity_type, 'node' AS field_revision_field_position_app_deadline_node_entity_type, 'node' AS field_revision_field_position_start_date_node_entity_type, 'node' AS field_revision_body_node_entity_type FROM node_revision node_revision LEFT JOIN node node_node_revision ON node_revision.nid = node_node_revision.nid LEFT JOIN field_revision_field_position_institution_ref field_revision_field_position_institution_ref ON node_revision.vid = field_revision_field_position_institution_ref.revision_id AND (field_revision_field_position_institution_ref.entity_type = 'node' AND field_revision_field_position_institution_ref.deleted = '0') LEFT JOIN node node_field_revision_field_position_institution_ref ON field_revision_field_position_institution_ref.field_position_institution_ref_target_id = node_field_revision_field_position_institution_ref.nid LEFT JOIN field_revision_field_position_cip_code field_revision_field_position_cip_code ON node_revision.vid = field_revision_field_position_cip_code.revision_id AND (field_revision_field_position_cip_code.entity_type = 'node' AND field_revision_field_position_cip_code.deleted = '0') LEFT JOIN node node_field_revision_field_position_cip_code ON field_revision_field_position_cip_code.field_position_cip_code_target_id = node_field_revision_field_position_cip_code.nid LEFT JOIN node node_node_revision_1 ON node_revision.nid = node_node_revision_1.nid LEFT JOIN field_revision_field_position_vacancy_status field_revision_field_position_vacancy_status ON node_revision.vid = field_revision_field_position_vacancy_status.revision_id AND (field_revision_field_position_vacancy_status.entity_type = 'node' AND field_revision_field_position_vacancy_status.deleted = '0') LEFT JOIN search_index search_index ON node_revision.nid = search_index.sid LEFT JOIN search_total search_total ON search_index.word = search_total.word WHERE ( ( (node_node_revision.status = '1') AND (node_node_revision.type IN ('position')) AND (field_revision_field_position_vacancy_status.field_position_vacancy_status_target_id IN ('38')) AND( (search_index.type = 'node') AND( (search_index.word = 'accountant') ) ) AND ( (node_revision.vid=node_node_revision.vid AND node_node_revision.status=1) ) ) ) GROUP BY search_index.sid, vid, score, field_data_field_system_inst_name_node_entity_type, field_revision_field_position_college_division_node_entity_t, field_revision_field_position_department_node_entity_type, field_revision_field_search_lvl_degree_lvls_node_entity_type, field_revision_field_position_app_deadline_node_entity_type, field_revision_field_position_start_date_node_entity_type, field_revision_body_node_entity_type HAVING ( ( (COUNT(*) >= '1') ) ) ORDER BY node_node_revision_title ASC LIMIT 20 OFFSET 0; Again, this query returns different sets of results from MySQL 5.5.28 (correct) to 5.6.14 (incorrect). If I move the column named "score" (the SUM() column) to the end of the column list, the query returns the correct set of results in both versions of MySQL. My question is: Is this expected behavior (and why), or is this a bug? I'm on the verge of reverting my entire environment back to 5.5 because of this.

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  • MySQL config for 2GB ram

    - by Tiffany Walker
    How is my config? Does it work well for 2GB? What would be an ideal config for a 2GB ram server? [mysqld] set-variable = max_connections=500 log-slow-queries safe-show-database local-infile=0 skip-networking symbolic-links=0 max_connections = 500 key_buffer = 256M myisam_sort_buffer_size = 64M join_buffer_size = 2M read_buffer_size = 2M sort_buffer_size = 2M read_rnd_buffer_size = 2M thread_concurrency = 16 table_cache = 1024 thread_cache_size = 50 wait_timeout = 7200 connect_timeout = 10 tmp_table_size = 32M max_allowed_packet = 160M max_connect_errors = 10 query_cache_limit = 1M query_cache_size = 32M query_cache_type = 1 [mysqld_safe] open_files_limit = 8192 [mysqldump] max_allowed_packet = 16M [myisamchk] key_buffer = 64M sort_buffer = 64M read_buffer = 16M write_buffer = 16M UPDATE 2012-03-28 12:58 EDT By RolandoMySQLDBA Please run these queries and paste them into your question: For MyISAM SELECT CONCAT(ROUND(KBS/POWER(1024, IF(PowerOf1024<0,0,IF(PowerOf1024>3,0,PowerOf1024)))+0.4999), SUBSTR(' KMG',IF(PowerOf1024<0,0, IF(PowerOf1024>3,0,PowerOf1024))+1,1)) recommended_key_buffer_size FROM (SELECT LEAST(POWER(2,32),KBS1) KBS FROM (SELECT SUM(index_length) KBS1 FROM information_schema.tables WHERE engine='MyISAM' AND table_schema NOT IN ('information_schema','mysql')) AA ) A, (SELECT 2 PowerOf1024) B; For InnoDB SELECT CONCAT(ROUND(KBS/POWER(1024, IF(PowerOf1024<0,0,IF(PowerOf1024>3,0,PowerOf1024)))+0.49999), SUBSTR(' KMG',IF(PowerOf1024<0,0, IF(PowerOf1024>3,0,PowerOf1024))+1,1)) recommended_innodb_buffer_pool_size FROM (SELECT SUM(data_length+index_length) KBS FROM information_schema.tables WHERE engine='InnoDB') A, (SELECT 2 PowerOf1024) B;

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  • What I should know about memory management?

    - by bua
    first of all: I don't use stackadmin or similar so please don't vote for moving there, I'm reading man top and paper "what every programmer should know about memory ..." I need really simple explanation like for retard ;) Having following top dump: top - 11:21:19 up 37 days, 21:16, 4 users, load average: 0.41, 0.75, 1.09 Tasks: 313 total, 5 running, 308 sleeping, 0 stopped, 0 zombie Cpu(s): 0.4%us, 0.6%sy, 0.9%ni, 96.2%id, 0.1%wa, 0.0%hi, 1.9%si, 0.0%st Mem: 132103848k total, 131916948k used, 186900k free, 54000k buffers Swap: 73400944k total, 73070884k used, 330060k free, 13931192k cached PID USER PR NI VIRT RES SHR S %CPU %MEM TIME+ COMMAND 3305 tudb 25 10 144m 52m 940 R 6.0 0.0 1306:09 app 3011 tudb 15 0 71528 19m 604 S 3.3 0.0 171:57.83 app 3373 tudb 25 10 209m 93m 940 S 3.0 0.1 1074:53 app 3338 tudb 25 10 144m 47m 940 R 2.7 0.0 780:48.48 app 4227 tudb 25 10 208m 99m 904 S 1.3 0.1 198:56.01 app 8506 tudb 25 10 80.7g 49g 932 S 2.0 39.6 458:31.22 app I'm wondering what is: RES (my expl. physical memory consumption ? see 49GB) VIRT (memory mapped disk to cache? see 80GB) SHR (shared pages?) Swap: (is this cached label - for memory mapped disk into swap cache?) Should sum of RES give MEM: X used? or maybe sum of VIRT?

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  • Python 3.4 adds re.fullmatch()

    - by Jan Goyvaerts
    Python 3.4 does not bring any changes to its regular expression syntax compared to previous 3.x releases. It does add one new function to the re module called fullmatch(). This function takes a regular expression and a subject string as its parameters. It returns True if the regular expression can match the string entirely. It returns False if the string cannot be matched or if it can only be matched partially. This is useful when using a regular expression to validate user input. Do note that fullmatch() will return True if the subject string is the empty string and the regular expression can find zero-length matches. A zero-length match of a zero-length string is a complete match. So if you want to check whether the user entered a sequence of digits, use \d+ rather than \d* as the regex.

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  • Project Euler 8: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 8.  As always, any feedback is welcome. # Euler 8 # http://projecteuler.net/index.php?section=problems&id=8 # Find the greatest product of five consecutive digits # in the following 1000-digit number import time start = time.time() number = '\ 73167176531330624919225119674426574742355349194934\ 96983520312774506326239578318016984801869478851843\ 85861560789112949495459501737958331952853208805511\ 12540698747158523863050715693290963295227443043557\ 66896648950445244523161731856403098711121722383113\ 62229893423380308135336276614282806444486645238749\ 30358907296290491560440772390713810515859307960866\ 70172427121883998797908792274921901699720888093776\ 65727333001053367881220235421809751254540594752243\ 52584907711670556013604839586446706324415722155397\ 53697817977846174064955149290862569321978468622482\ 83972241375657056057490261407972968652414535100474\ 82166370484403199890008895243450658541227588666881\ 16427171479924442928230863465674813919123162824586\ 17866458359124566529476545682848912883142607690042\ 24219022671055626321111109370544217506941658960408\ 07198403850962455444362981230987879927244284909188\ 84580156166097919133875499200524063689912560717606\ 05886116467109405077541002256983155200055935729725\ 71636269561882670428252483600823257530420752963450' max = 0 for i in xrange(0, len(number) - 5): nums = [int(x) for x in number[i:i+5]] val = reduce(lambda agg, x: agg*x, nums) if val > max: max = val print max print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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  • Why is it impossible to produce truly random numbers?

    - by Vinoth Kumar
    I was trying to solve a hobby problem that required generating a million random numbers. But I quickly realized, it is becoming difficult to make them unique. I picked up Algorithm Design Manual to read about random number generation. It has the following paragraph that I am fully not able to understand. Unfortunately, generating random numbers looks a lot easier than it really is. Indeed, it is fundamentally impossible to produce truly random numbers on any deterministic device. Von Neumann [Neu63] said it best: “Anyone who considers arithmetical methods of producing random digits is, of course, in a state of sin.” The best we can hope for are pseudo-random numbers, a stream of numbers that appear as if they were generated randomly. Why is it impossible to produce truly random numbers in any deterministic device? What does this sentence mean?

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  • Using a Predicate as a key to a Dictionary

    - by Tom Hines
    I really love Linq and Lambda Expressions in C#.  I also love certain community forums and programming websites like DaniWeb. A user on DaniWeb posted a question about comparing the results of a game that is like poker (5-card stud), but is played with dice. The question stemmed around determining what was the winning hand.  I looked at the question and issued some comments and suggestions toward a potential answer, but I thought it was a neat homework exercise. [A little explanation] I eventually realized not only could I compare the results of the hands (by name) with a certain construct – I could also compare the values of the individual dice with the same construct. That piece of code eventually became a Dictionary with the KEY as a Predicate<int> and the Value a Func<T> that returns a string from the another structure that contains the mapping of an ENUM to a string.  In one instance, that string is the name of the hand and in another instance, it is a string (CSV) representation of of the digits in the hand. An added benefit is that the digits re returned in the order they would be for a proper poker hand.  For instance the hand 1,2,5,3,1 would be returned as ONE_PAIR (1,1,5,3,2). [Getting to the point] 1: using System; 2: using System.Collections.Generic; 3:   4: namespace DicePoker 5: { 6: using KVP_E2S = KeyValuePair<CDicePoker.E_DICE_POKER_HAND_VAL, string>; 7: public partial class CDicePoker 8: { 9: /// <summary> 10: /// Magical construction to determine the winner of given hand Key/Value. 11: /// </summary> 12: private static Dictionary<Predicate<int>, Func<List<KVP_E2S>, string>> 13: map_prd2fn = new Dictionary<Predicate<int>, Func<List<KVP_E2S>, string>> 14: { 15: {new Predicate<int>(i => i.Equals(0)), PlayerTie},//first tie 16:   17: {new Predicate<int>(i => i > 0), 18: (m => string.Format("Player One wins\n1={0}({1})\n2={2}({3})", 19: m[0].Key, m[0].Value, m[1].Key, m[1].Value))}, 20:   21: {new Predicate<int>(i => i < 0), 22: (m => string.Format("Player Two wins\n2={2}({3})\n1={0}({1})", 23: m[0].Key, m[0].Value, m[1].Key, m[1].Value))}, 24:   25: {new Predicate<int>(i => i.Equals(0)), 26: (m => string.Format("Tie({0}) \n1={1}\n2={2}", 27: m[0].Key, m[0].Value, m[1].Value))} 28: }; 29: } 30: } When this is called, the code calls the Invoke method of the predicate to return a bool.  The first on matching true will have its value invoked. 1: private static Func<DICE_HAND, E_DICE_POKER_HAND_VAL> GetHandEval = dh => 2: map_dph2fn[map_dph2fn.Keys.Where(enm2fn => enm2fn(dh)).First()]; After coming up with this process, I realized (with a little modification) it could be called to evaluate the individual values in the dice hand in the event of a tie. 1: private static Func<List<KVP_E2S>, string> PlayerTie = lst_kvp => 2: map_prd2fn.Skip(1) 3: .Where(x => x.Key.Invoke(RenderDigits(dhPlayerOne).CompareTo(RenderDigits(dhPlayerTwo)))) 4: .Select(s => s.Value) 5: .First().Invoke(lst_kvp); After that, I realized I could now create a program completely without “if” statements or “for” loops! 1: static void Main(string[] args) 2: { 3: Dictionary<Predicate<int>, Action<Action<string>>> main = new Dictionary<Predicate<int>, Action<Action<string>>> 4: { 5: {(i => i.Equals(0)), PlayGame}, 6: {(i => true), Usage} 7: }; 8:   9: main[main.Keys.Where(m => m.Invoke(args.Length)).First()].Invoke(Display); 10: } …and there you have it. :) ZIPPED Project

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  • How do I get a Netgear WNDA3100V2 working?

    - by Michal
    I have Ubuntu 11.10 on my desktop. A month ago I bought Linksys AE1000 adapter,I did not check that it's not working on Ubuntu and because I've lost receipt I'm stuck with it. Last week I bought Netgear adapter and this time I did check and it meant to be plug an play but it was not. I have checked many forums and managed to install software, system does sees adapter but it's not connecting to network. I have found that it may not like WPA so I have created my own password-letters and digits,no spaces-still nothing.I don't understand why. This is my next attempt with Linux and I'm not with IT background so it takes time and research before I can resolve something but I really want to learn. I so wish to learn on Ubuntu.One day, I've checked Fedora16 and my old Linksys AE1000 worked without any instalations.

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  • Perl numerical sorting: how to ignore leading alpha character [migrated]

    - by Luke Sheppard
    I have a 1,660 row array like this: ... H00504 H00085 H00181 H00500 H00103 H00007 H00890 H08793 H94316 H00217 ... And the leading character never changes. It is always "H" then five digits. But when I do what I believe is a numerical sort in Perl, I'm getting strange results. Some segments are sorted in order, but then a different segment starts up. Here is a segment after sorting: ... H01578 H01579 H01580 H01581 H01582 H01583 H01584 H00536 H00537 H00538 H01585 H01586 H01587 H01588 H01589 H01590 ... What I'm trying is this: my @sorted_array = sort {$a <=> $b} @raw_array; But obviously it is not working. Anyone know why?

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  • Project Euler 13: (Iron)Python

    - by Ben Griswold
    In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 13.  As always, any feedback is welcome. # Euler 13 # http://projecteuler.net/index.php?section=problems&id=13 # Work out the first ten digits of the sum of the # following one-hundred 50-digit numbers. import time start = time.time() number_string = '\ 37107287533902102798797998220837590246510135740250\ 46376937677490009712648124896970078050417018260538\ 74324986199524741059474233309513058123726617309629\ 91942213363574161572522430563301811072406154908250\ 23067588207539346171171980310421047513778063246676\ 89261670696623633820136378418383684178734361726757\ 28112879812849979408065481931592621691275889832738\ 44274228917432520321923589422876796487670272189318\ 47451445736001306439091167216856844588711603153276\ 70386486105843025439939619828917593665686757934951\ 62176457141856560629502157223196586755079324193331\ 64906352462741904929101432445813822663347944758178\ 92575867718337217661963751590579239728245598838407\ 58203565325359399008402633568948830189458628227828\ 80181199384826282014278194139940567587151170094390\ 35398664372827112653829987240784473053190104293586\ 86515506006295864861532075273371959191420517255829\ 71693888707715466499115593487603532921714970056938\ 54370070576826684624621495650076471787294438377604\ 53282654108756828443191190634694037855217779295145\ 36123272525000296071075082563815656710885258350721\ 45876576172410976447339110607218265236877223636045\ 17423706905851860660448207621209813287860733969412\ 81142660418086830619328460811191061556940512689692\ 51934325451728388641918047049293215058642563049483\ 62467221648435076201727918039944693004732956340691\ 15732444386908125794514089057706229429197107928209\ 55037687525678773091862540744969844508330393682126\ 18336384825330154686196124348767681297534375946515\ 80386287592878490201521685554828717201219257766954\ 78182833757993103614740356856449095527097864797581\ 16726320100436897842553539920931837441497806860984\ 48403098129077791799088218795327364475675590848030\ 87086987551392711854517078544161852424320693150332\ 59959406895756536782107074926966537676326235447210\ 69793950679652694742597709739166693763042633987085\ 41052684708299085211399427365734116182760315001271\ 65378607361501080857009149939512557028198746004375\ 35829035317434717326932123578154982629742552737307\ 94953759765105305946966067683156574377167401875275\ 88902802571733229619176668713819931811048770190271\ 25267680276078003013678680992525463401061632866526\ 36270218540497705585629946580636237993140746255962\ 24074486908231174977792365466257246923322810917141\ 91430288197103288597806669760892938638285025333403\ 34413065578016127815921815005561868836468420090470\ 23053081172816430487623791969842487255036638784583\ 11487696932154902810424020138335124462181441773470\ 63783299490636259666498587618221225225512486764533\ 67720186971698544312419572409913959008952310058822\ 95548255300263520781532296796249481641953868218774\ 76085327132285723110424803456124867697064507995236\ 37774242535411291684276865538926205024910326572967\ 23701913275725675285653248258265463092207058596522\ 29798860272258331913126375147341994889534765745501\ 18495701454879288984856827726077713721403798879715\ 38298203783031473527721580348144513491373226651381\ 34829543829199918180278916522431027392251122869539\ 40957953066405232632538044100059654939159879593635\ 29746152185502371307642255121183693803580388584903\ 41698116222072977186158236678424689157993532961922\ 62467957194401269043877107275048102390895523597457\ 23189706772547915061505504953922979530901129967519\ 86188088225875314529584099251203829009407770775672\ 11306739708304724483816533873502340845647058077308\ 82959174767140363198008187129011875491310547126581\ 97623331044818386269515456334926366572897563400500\ 42846280183517070527831839425882145521227251250327\ 55121603546981200581762165212827652751691296897789\ 32238195734329339946437501907836945765883352399886\ 75506164965184775180738168837861091527357929701337\ 62177842752192623401942399639168044983993173312731\ 32924185707147349566916674687634660915035914677504\ 99518671430235219628894890102423325116913619626622\ 73267460800591547471830798392868535206946944540724\ 76841822524674417161514036427982273348055556214818\ 97142617910342598647204516893989422179826088076852\ 87783646182799346313767754307809363333018982642090\ 10848802521674670883215120185883543223812876952786\ 71329612474782464538636993009049310363619763878039\ 62184073572399794223406235393808339651327408011116\ 66627891981488087797941876876144230030984490851411\ 60661826293682836764744779239180335110989069790714\ 85786944089552990653640447425576083659976645795096\ 66024396409905389607120198219976047599490197230297\ 64913982680032973156037120041377903785566085089252\ 16730939319872750275468906903707539413042652315011\ 94809377245048795150954100921645863754710598436791\ 78639167021187492431995700641917969777599028300699\ 15368713711936614952811305876380278410754449733078\ 40789923115535562561142322423255033685442488917353\ 44889911501440648020369068063960672322193204149535\ 41503128880339536053299340368006977710650566631954\ 81234880673210146739058568557934581403627822703280\ 82616570773948327592232845941706525094512325230608\ 22918802058777319719839450180888072429661980811197\ 77158542502016545090413245809786882778948721859617\ 72107838435069186155435662884062257473692284509516\ 20849603980134001723930671666823555245252804609722\ 53503534226472524250874054075591789781264330331690' total = 0 for i in xrange(0, 100 * 50 - 1, 50): total += int(number_string[i:i+49]) print str(total)[:10] print "Elapsed Time:", (time.time() - start) * 1000, "millisecs" a=raw_input('Press return to continue')

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