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  • Radius of multiple latitude/longitude points

    - by zekial
    I have a program that takes as input an array of lat/long points. I need to perform a check on that array to ensure that all of the points are within a certain radius. So, for example, the maximum radius I will allow is 100 miles. Given an array of lat/long (coming from a MySQL database, could be 10 points could be 10000) I need to figure out if they will all fit in a circle with radius of 100 miles. Kinda stumped on how to approach this. Any help would be greatly appreciated.

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  • Code Golf: All +-*/ Combinations for 3 integers

    - by Flash84x
    Write a program that takes 3 integers separated by spaces and perform every single combination of addition, subtraction, multiplication and division operations possible and display the result with the operation combination used. Example: $./solution 1 2 3 Results in the following output 1+2+3 = 6 1-2-3 = -4 1*2*3 = 6 1/2/3 = 0 (integer answers only, round up at .5) 1*2-3 = -1 3*1+2 = 5 etc... Order of operation rules apply, assume there will be no parenthesis used i.e. (3-1)*2 = 4 is not a combination, although you could implement this for "extra credit" For results where a divide by 0 occurs simply return NaN

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  • Multiple left joins, how to output in php

    - by Dan
    I have 3 tables I need to join. The contracts table is the main table, the 'jobs' and 'companies' table are extra info that can be associated to the contracts table. so, since I want all entries from my 'contracts' table, and the 'jobs' and 'companies' data only if it exists, I wrote the query like this.... $sql = "SELECT * FROM contracts LEFT JOIN jobs ON contracts.job_id = jobs.id LEFT JOIN companies ON contracts.company_id = companies.id ORDER BY contracts.end_date"; Now how would I output this in PHP? I tried this but kept getting an undefined error "Notice: Undefined index: contracts.id"... $sql_result = mysql_query($sql,$connection) or die ("Fail."); if(mysql_num_rows($sql_result) > 0){ while($row = mysql_fetch_array($sql_result)) { $contract_id = stripslashes($row['contracts.id']); $job_number = stripslashes($row['jobs.job_number']); $company_name = stripslashes($row['companies.name']); ?> <tr id="<?=$contract_id?>"> <td><?=$job_number?></td> <td><?=$company_name?></td> </tr> <? } }else{ echo "No records found"; } Any help is appreciated.

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  • Subdividing 3D mesh into arbitrarily sized pieces

    - by Groky
    I have a mesh defined by 4 points in 3D space. I need an algorithm which will subdivide that mesh into subdivisions of an arbitrary horizontal and vertical size. If the subdivision size isn't an exact divisor of the mesh size, the edge pieces will be smaller. All of the subdivision algorithms I've found only subdivide meshes into exact powers of 2. Does anyone know of one that can do what I want? Failing that, my thoughts about a possible implementation is to rotate the mesh so that it is flat on the Z axis, subdivide in 2D and then translate back into 3D. That's because my mind finds 3D hard ;) Any better suggestions? Using C# if that makes any difference.

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  • How to detect if a certain range resides (partly) within an other range?

    - by Tom
    Lets say I've got two squares and I know their positions, a red and blue square: redTopX; redTopY; redBotX; redBotY; blueTopX; blueTopY; blueBotX; blueBotY; Now, I want to check if square blue resides (partly) within (or around) square red. This can happen in a lot of situations, as you can see in this image I created to illustrate my situation better: Note that there's always only one blue and one red square, I just added multiple so I didn't have to redraw 18 times. My original logic was simple, I'd check all corners of square blue and see if any of them are inside square red: if ( ((redTopX >= blueTopX) && (redTopY >= blueTopY) && (redTopX <= blueBotX) && (redTopY <= blueBotY)) || //top left ((redBotX >= blueTopX) && (redTopY >= blueTopY) && (redBotX <= blueBotX) && (redTopY <= blueBotY)) || //top right ((redTopX >= blueTopX) && (redBotY >= blueTopY) && (redTopX <= blueBotX) && (redBotY <= blueBotY)) || //bottom left ((redBotX >= blueTopX) && (redBotY >= blueTopY) && (redBotX <= blueBotX) && (redBotY <= blueBotY)) //bottom right ) { //blue resides in red } Unfortunately, there are a couple of flaws in this logic. For example, what if red surrounds blue (like in situation 1)? I thought this would be pretty easy but am having trouble coming up with a good way of covering all these situations.. can anyone help me out here? Regards, Tom

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  • Fastest way to find the rotation of a vector

    - by kriss
    I have two 2D vectors, say u and v, defined by cartesian coordinates. Imagine that vectors are needles of a clock. I'm looking for the fastest way to find out, using python, if v is after or before u (or in other words find out in wich half plane is v, regarding to position of u). For the purpose of the problem if vectors are aligned answer should be before. It seems easy using some trigonometry, but I believe there should be a faster way using coordinates only. My test case: def after(u, v): """code here""" after((4,2), (6, 1)) : True after((4,2), (3, 3)) : False after((4,2), (2, 1)) : False after((4,2), (3, -3)) : True after((4,2), (-2, -5)) : True after((4,2), (-4, -2)) : False

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  • All non-prime factorings

    - by Tony Veijalainen
    Let's say we have numbers factors, for example 1260: >>> factors(1260) [2, 2, 3, 3, 5, 7] Which would be best way to do in Python combinations with every subproduct possible from these numbers, ie all factorings, not only prime factoring, with sum of factors less than max_sum? If I do combinations from the prime factors, I have to refactor remaining part of the product as I do not know the remaining part not in combination. Example results would be: [4, 3, 3, 5, 7] (one replacement) [3, 6, 14, 5] (two replacements)

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  • Shortest command to calculate the sum of a column of output on Unix?

    - by Andrew
    I'm sure there is a quick and easy way to calculate the sum of a column of values on Unix systems (using something like awk or xargs perhaps), but writing a shell script to parse the rows line by line is the only thing that comes to mind at the moment. For example, what's the simplest way to modify the command below to compute and display the total for the SEGSZ column (70300)? ipcs -mb | head -6 IPC status from /dev/kmem as of Mon Nov 17 08:58:17 2008 T ID KEY MODE OWNER GROUP SEGSZ Shared Memory: m 0 0x411c322e --rw-rw-rw- root root 348 m 1 0x4e0c0002 --rw-rw-rw- root root 61760 m 2 0x412013f5 --rw-rw-rw- root root 8192

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  • How to fix this 7% where 6% , 5% works great.

    - by Stackfan
    Case 1 (discount 6%): Subtotal: 750.00 Discount: 45.00 Handling cost: 24.32 21% VAT: 0.00 Total (this is the amount you will deposit): 729.32 Case 2 (discount 7%): Subtotal: 1250.00 Discount: 87.50 Handling cost: 39.88 21% VAT: 0.00 Total (this is the amount you will deposit): 1202.38 Where i am applying this formula: (729.32 - 0.35) / 1.034/ 0.94 = 750.00 (<<--- CORRECT ) ? (1202.38 - 0.35) / 1.034/ 0.93 = 1250.01 (<<--- My problem why not 1250.00) ? How to correct the 7% formula to get exactly 1250.00 ? Instead of fraction error.

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  • getting the value of a filter at an arbitrary time

    - by Andiih
    Context: I'm trying to improve the values returned by the iPhone CLLocationManager, although this is a more generally applicable problem. The key is that CLLocationManger returns data on current velocity as and when it feels like it, rather than at a fixed sample rate. I'd like to use a feedback equation to improve accuracy v=(k*v)+(1-k)*currentVelocity where currentVelocity is the speed returned by didUpdateToLocation:fromLocation: and v is the output velocity (and also used for the feedback element). Because of the "as and when" nature of didUpdateToLocation:fromLocation: I could calculate the time interval since it was last called, and do something like for (i=0;i<timeintervalsincelastcalled;i++) v=(k*v)+(1-k)*currentVelocity which would work, but is wasteful of cycles. Especially as I probably want timeintervalsincelastcalled to be measured as 10ths of a second. Is there a way to solve this without the loop ? i.e. rework (integrate?) the formula so I put an interval into the equation and get the same answer as I would have by iteration ?

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  • Faster way to compare two sets of points in N-dimensional space?

    - by Amit
    List1 contains a high number (~7^10) of N-dimensional points (N <=10), List2 contains the same or fewer number of N-dimensional points (N <=10). My task is this: I want to check which point in List2 is closest (euclidean distance) to a point in List1 for every point in List1 and subsequently perform some operation on it. I have been doing it the simple- the nested loop way when I didn't have more than 50 points in List1, but with 7^10 points, this obviously takes up a lot of time. What is the fastest way to do this? Any concepts from Computational Geometry might help?

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  • 6^x = 5 equation, how to solve it?

    - by Tom
    If it would be 6^x = 1 or 6^x = 6 or 6^x = 36 it would be extremely easy, but how to solve this equation: 6^x = 5 I don't need an answer, I want to find out how to solve equations like this one, I need solution. Thanks.

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  • simple rank formula

    - by Graham
    I'm looking for a mathmatical ranking formula. Sample is 2008 2009 2010 A 5 6 4 B 6 7 5 C 7 8 2 I want to add a rank column for each period code field rank 2008 2009 2010 2008 2009 2010 B 6 7 5 2 1 1 A 5 6 4 3 2 2 C 7 2 2 1 3 3 please do not reply with methods that loop thru the rows and columns, incrementing the rank value as it goes, that's easy. I'm looking for a formula much like finding the percent total (item / total). I know i've seen this before but an havning a tough time locating it. Thanks in advance!

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  • Drawing Directed Acyclic Graphs: Using DAG property to improve layout/edge routing?

    - by Robert Fraser
    Hi, Laying out the verticies in a DAG in a tree form (i.e. verticies with no in-edges on top, verticies dependent only on those on the next level, etc.) is rather simple. However, is there a simple algorithm to do this that minimizes edge crossing? (For some graphs, it may be impossible to completely eliminate edge crossing.) A picture says a thousand words, so is there an algorithm that would suggest: instead of:

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  • Distributing players to tables

    - by IVlad
    Consider N = 4k players, k tables and a number of clans such that each member can belong to one clan. A clan can contain at most k players. We want to organize 3 rounds of a game such that, for each table that seats exactly 4 players, no 2 players sitting there are part of the same clan, and, for the later rounds, no 2 players sitting there have sat at the same table before. All players play all rounds. How can we do this efficiently if N can be about ~80 large? I thought of this: for each table T: repeat until 4 players have been seated at T: pick a random player X that is not currently seated anywhere if X has not sat at the same table as anyone currently at T AND X is not from the same clan as anyone currently at T seat X at T break I am not sure if this will always finish or if it can get stuck even if there is a valid assignment. Even if this works, is there a better way to do it?

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  • Android: Find coordinates of a certain point X meters from my location moving towards the point I am

    - by Aidan
    Hi Guys, I'm constructing a geolocation based application and I'm trying to figure out a way to make my application realise when a user is facing the direction of the given location (a particular long / lat co-ord). I've done some Googling and checked the SDK but can't really find anything for such a thing. Does anyone know of a way? Example. Point A = Phones current location. Point B = A's orientation in relation to true north + 45 + max distance towards the direction your facing, Point C = A's orientation in relation to true north - 45 + max distance towards the direction your facing. So now you have a triangle constructed. pretty schweet huh? yeah.. I think so.. So now that I have my fancy Triangle I use something called Barycentric Coordinates ( http://en.wikipedia.org/wiki/Barycentric_coordinates_(mathematics) ). This will allow me to test another point and see if it is in the triangle. If it is, it means we're facing it AND it's within the right distance. So it should be displayed on screen. If I'm facing 90 degrees from true north. The distance it travels should be that direction. 90 degrees from true north. It should not be 100 degrees or something from true north! But the problem is I haven't yet figured out how I make the device realise it must go "out" the direction it is facing.

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  • Linear Interpolation. How to implement this algorithm in C ? (Python version is given)

    - by psihodelia
    There exists one very good linear interpolation method. It performs linear interpolation requiring at most one multiply per output sample. I found its description in a third edition of Understanding DSP by Lyons. This method involves a special hold buffer. Given a number of samples to be inserted between any two input samples, it produces output points using linear interpolation. Here, I have rewritten this algorithm using Python: temp1, temp2 = 0, 0 iL = 1.0 / L for i in x: hold = [i-temp1] * L temp1 = i for j in hold: temp2 += j y.append(temp2 *iL) where x contains input samples, L is a number of points to be inserted, y will contain output samples. My question is how to implement such algorithm in ANSI C in a most effective way, e.g. is it possible to avoid the second loop? NOTE: presented Python code is just to understand how this algorithm works. UPDATE: here is an example how it works in Python: x=[] y=[] hold=[] num_points=20 points_inbetween = 2 temp1,temp2=0,0 for i in range(num_points): x.append( sin(i*2.0*pi * 0.1) ) L = points_inbetween iL = 1.0/L for i in x: hold = [i-temp1] * L temp1 = i for j in hold: temp2 += j y.append(temp2 * iL) Let's say x=[.... 10, 20, 30 ....]. Then, if L=1, it will produce [... 10, 15, 20, 25, 30 ...]

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