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  • How do you react to non-programmers with ideas of 'The Next Big Thing' ?

    - by jiceo
    Recently and quite often, people with no programming background come and say they have this great idea that can be the next big thing and that the idea(s) is worth a fortune by itself. Then as they know I'm a programmer, they ask me if I'm willing to "code it up" for them or find someone willing to do it for next to nothing. Judging from the enthusiasm, it's like they're drunk on their idea and that that by itself is the most important thing, but they just need a programmer. My response to them, depending on my mood and their general attitude towards what we do, is something along the lines of: "Having the core of an idea is one thing. Developing it to the point that it becomes a platform that changes the world in which it lives is another, and you're going to be willing to pay proportionately to how big you think your idea is worth." Have you been approached by these business type entrepreneurs (with no technical/developer's knowledge) with such a proposal and how do you react to them?

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  • I prefer C/C++ over Unity and other tools: is it such a big downer for a game developer?

    - by jokoon
    We have a big game project using Unity at school. There are 12 of us working on it. My teacher seems to be convinced it's an important tool to teach students, since it makes students look from the high level to the lower level. I can understand his view, and I'm wondering: Is unity such an important engine in game development companies? Are there a lot of companies using it because they can't afford to use something else? He is talking like Unity is a big player in game making, but I only see it fit small indie game companies who want to do a game as fast as possible. Do you think Unity is of that much importance in the industry? Does it endanger the value of C++ skills? It's not that I don't like Unity, it's just that I don't learn anything with it, I prefer to achieve little steps with Ogre or SFML instead. Also, we also have C++ practice exercises, but those are just practice with theory, nothing much.

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  • What are some concepts people should understand before programming "big" projects?

    - by Abafei
    A person new to programming may be able to make a good small program. However, when starting to work on anything bigger than a small (think 1 C source file or Python module) program, there are some general concepts which become much more important when working on "big" (think many Python modules or C files) programs; one example is modularity, another is having a set aim. Some of these may be obvious to people who went to school to learn programming; however, people like me who did not go to programming classes sometimes have to learn these things from experience, possibly creating failed projects in the meantime. ================================================== Please explain what the concept is, and why the concept becomes more important for big programs than by small programs. Please give only 1 concept per answer.

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  • IBM : "les ordinateurs pourront voir, sentir, toucher, gouter et entendre" d'ici 5 ans, Big Blue livre ses prédictions "5 in 5"

    IBM : « les systèmes informatiques pourront voir, sentir, toucher, gouter et entendre » d'ici 5 ans Big Blue livre ses prédictions « Five in Five » Comme il est de coutume en chaque fin d'année, IBM vient de livrer ses cinq prédictions sur l'évolution de la technologie au cours des cinq années à venir. Big Blue lors de son événement « Five In Five » a publié sa vision d'un futur ou les dispositifs informatiques seront dotés des cinq sens. Ils seront capables de voir, sentir, toucher, gouter et entendre. Le toucher : un téléphone sera capable de reproduire une sensation du toucher De nos jours, les technologies haptiques et graphiques utilisées dans le domaine ...

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  • Split a big folder (and its contents) into several small new folders ?

    - by David
    Hello, By luck do you know a software on Windows (Vista) which can easily split a big folder (and its contents) into several small new folders ? notes: the new small folders should be independent the "file splitter" that I have already tested don't do alas the "folder splitter" job ! ;( Example : I have a main big folder C:\big\ (size: about 3Go) The result I need : C:\big01\ (size: about 200MB) C:\big02\ (size: about 200MB) etc... Thanks in advance ;)

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  • WCF Service behind F5 Bigip endpoint issue

    - by John
    we have WCF services hosted on IIS in 4 web servers, these web servers are behind the F5 Bigip load balancing system, SSL accelarator. When the client calls the service it'll be calling https myserver.com/myservices.svc but actually .svc will be in different servers like http web1.mysever.com/myservices.svc, http web2.mysever.com/myservices.svc, etc how do we handle this issue?

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  • WCF, Metadata and BIGIP - Can I force the correct url for the WSDL items?

    - by Yossi Dahan
    We have a WCF service hosted on ServerA which is a server with no-direct Internet access and has a non-Internet routable IP address. The service is fronted by BIGIP which handles SSL encryption and decryption and forwards the unencrypted request to ServerA (at the moment it does NOT actually do any load balancing, but that is likely to be added in the future) on a specific port. What that means is that our clients would be calling the service through https://www.OurDomain.com/ServiceUrl and would get to our service on http://SeverA:85/ServiceUrl through the BIGIP device; When we browse to the WSDL published on https://www.OurDomain.com/ServiceUrl all the addresses contained in the WSDL are based on the http://SeverA:85/ServiceUrl base address We figured out that we could use the host headers setting to set the domain, but our problem is that while this would sort out the domain, we would still be using the wrong scheme – it would use http://www.OurDomain.com/ServiceUrl while we need it to be Https. Also – as we have other services (asmx based) hosted on that server we had some issues setting the host headers, and so we thought we could get away with creating another site on the server (using, say, port 82) and set the host header on that; now, on top of the http/https problem we have an issue as the WSDL contains the port number in all the urls, where BigIP works on port 443 (for the SSL) Is there a more flexible solution than implementing Host Headers? Ideally we need to retain flexibility and ease of supportability. Thanks for any help…

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  • Are there any worse sorting algorithms than Bogosort (a.k.a Monkey Sort)?

    - by womp
    My co-workers took me back in time to my University days with a discussion of sorting algorithms this morning. We reminisced about our favorites like StupidSort, and one of us was sure we had seen a sort algorithm that was O(n!). That got me started looking around for the "worst" sorting algorithms I could find. We postulated that a completely random sort would be pretty bad (i.e. randomize the elements - is it in order? no? randomize again), and I looked around and found out that it's apparently called BogoSort, or Monkey Sort, or sometimes just Random Sort. Monkey Sort appears to have a worst case performance of O(∞), a best case performance of O(n), and an average performance of O(n * n!). Are there any named algorithms that have worse average performance than O(n * n!)? Or are just sillier than Monkey Sort in general?

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  • Linear complexity and quadratic complexity

    - by jasonline
    I'm just not sure... If you have a code that can be executed in either of the following complexities: A sequence of O(n), like for example: two O(n) in sequence O(n²) The preferred version would be the one that can be executed in linear time. Would there be a time such that the sequence of O(n) would be too much and that O(n²) would be preferred? In other words, is the statement C x O(n) < O(n²) always true for any constant C? Why or why not? What are the factors that would affect the condition such that it would be better to choose the O(n²) complexity?

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  • Best way to do powerOf(int x, int n)?

    - by Mike
    So given x, and power, n, solve for X^n. There's the easy way that's O(n)... I can get it down to O(n/2), by doing numSquares = n/2; numOnes = n%2; return (numSquares * x * x + numOnes * x); Now there's a log(n) solution, does anyone know how to do it? It can be done recursively.

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  • How to analyze the efficiency of this algorithm Part 2

    - by Leonardo Lopez
    I found an error in the way I explained this question before, so here it goes again: FUNCTION SEEK(A,X) 1. FOUND = FALSE 2. K = 1 3. WHILE (NOT FOUND) AND (K < N) a. IF (A[K] = X THEN 1. FOUND = TRUE b. ELSE 1. K = K + 1 4. RETURN Analyzing this algorithm (pseudocode), I can count the number of steps it takes to finish, and analyze its efficiency in theta notation, T(n), a linear algorithm. OK. This following code depends on the inner formulas inside the loop in order to finish, the deal is that there is no variable N in the code, therefore the efficiency of this algorithm will always be the same since we're assigning the value of 1 to both A & B variables: 1. A = 1 2. B = 1 3. UNTIL (B > 100) a. B = 2A - 2 b. A = A + 3 Now I believe this algorithm performs in constant time, always. But how can I use Algebra in order to find out how many steps it takes to finish?

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  • How is schoolbook long division an O(n^2) algorithm?

    - by eSKay
    Premise: This Wikipedia page suggests that the computational complexity of Schoolbook long division is O(n^2). Deduction: Instead of taking "Two n-digit numbers", if I take one n-digit number and one m-digit number, then the complexity would be O(n*m). Contradiction: Suppose you divide 100000000 (n digits) by 1000 (m digits), you get 100000, which takes six steps to arrive at. Now, if you divide 100000000 (n digits) by 10000 (m digits), you get 10000 . Now this takes only five steps. Conclusion: So, it seems that the order of computation should be something like O(n/m). Question: Who is wrong, me or Wikipedia, and where?

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  • Asymptotic runtime of list-to-tree function

    - by Deestan
    I have a merge function which takes time O(log n) to combine two trees into one, and a listToTree function which converts an initial list of elements to singleton trees and repeatedly calls merge on each successive pair of trees until only one tree remains. Function signatures and relevant implementations are as follows: merge :: Tree a -> Tree a -> Tree a --// O(log n) where n is size of input trees singleton :: a -> Tree a --// O(1) empty :: Tree a --// O(1) listToTree :: [a] -> Tree a --// Supposedly O(n) listToTree = listToTreeR . (map singleton) listToTreeR :: [Tree a] -> Tree a listToTreeR [] = empty listToTreeR (x:[]) = x listToTreeR xs = listToTreeR (mergePairs xs) mergePairs :: [Tree a] -> [Tree a] mergePairs [] = [] mergePairs (x:[]) = [x] mergePairs (x:y:xs) = merge x y : mergePairs xs This is a slightly simplified version of exercise 3.3 in Purely Functional Data Structures by Chris Okasaki. According to the exercise, I shall now show that listToTree takes O(n) time. Which I can't. :-( There are trivially ceil(log n) recursive calls to listToTreeR, meaning ceil(log n) calls to mergePairs. The running time of mergePairs is dependent on the length of the list, and the sizes of the trees. The length of the list is 2^h-1, and the sizes of the trees are log(n/(2^h)), where h=log n is the first recursive step, and h=1 is the last recursive step. Each call to mergePairs thus takes time (2^h-1) * log(n/(2^h)) I'm having trouble taking this analysis any further. Can anyone give me a hint in the right direction?

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  • Unix: millionth number in the serie 2 3 4 6 9 13 19 28 42 63 ... ?

    - by HH
    It takes about minute to achieve 3000 in my comp but I need to know the millionth number in the serie. The definition is recursive so I cannot see any shortcuts except to calculate everything before the millionth number. How can you fast calculate millionth number in the serie? Serie Def n_{i+1} = \floor{ 3/2 * n_{i} } and n_{0}=2. Interestingly, only one site list the serie according to Goolge: this one. Too slow Bash code #!/bin/bash function serie { n=$( echo "3/2*$n" | bc -l | tr '\n' ' ' | sed -e 's@\\@@g' -e 's@ @@g' ); # bc gives \ at very large numbers, sed-tr for it n=$( echo $n/1 | bc ) #DUMMY FLOOR func } n=2 nth=1 while [ true ]; #$nth -lt 500 ]; do serie $n # n gets new value in the function throught global value echo $nth $n nth=$( echo $nth + 1 | bc ) #n++ done

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  • Find if there is an element repeating itself n/k times

    - by gleb-pendler
    You have an array size n and a constant k (whatever) You can assume the the array is of int type (although it could be of any type) Describe an algorithm that finds if there is an element(s) that repeats itself at least n/k times... if there is return one. Do so in linear time (O(n)) The catch: do this algorithm (or even pseudo-code) using constant memory and running over the array only twice

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  • minimum L sum in a mxn matrix - 2

    - by hilal
    Here is my first question about maximum L sum and here is different and hard version of it. Problem : Given a mxn *positive* integer matrix find the minimum L sum from 0th row to the m'th row . L(4 item) likes chess horse move Example : M = 3x3 0 1 2 1 3 2 4 2 1 Possible L moves are : (0 1 2 2), (0 1 3 2) (0 1 4 2) We should go from 0th row to the 3th row with minimum sum I solved this with dynamic-programming and here is my algorithm : 1. Take a mxn another Minimum L Moves Sum array and copy the first row of main matrix. I call it (MLMS) 2. start from first cell and look the up L moves and calculate it 3. insert it in MLMS if it is less than exists value 4. Do step 2. until m'th row 5. Choose the minimum sum in the m'th row Let me explain on my example step by step: M[ 0 ][ 0 ] sum(L1 = (0, 1, 2, 2)) = 5 ; sum(L2 = (0,1,3,2)) = 6; so MLMS[ 0 ][ 1 ] = 6 sum(L3 = (0, 1, 3, 2)) = 6 ; sum(L4 = (0,1,4,2)) = 7; so MLMS[ 2 ][ 1 ] = 6 M[ 0 ][ 1 ] sum(L5 = (1, 0, 1, 4)) = 6; sum(L6 = (1,3,2,4)) = 10; so MLMS[ 2 ][ 2 ] = 6 ... the last MSLS is : 0 1 2 4 3 6 6 6 6 Which means 6 is the minimum L sum that can be reach from 0 to the m. I think it is O(8*(m-1)*n) = O(m*n). Is there any optimal solution or dynamic-programming algorithms fit this problem? Thanks, sorry for long question

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  • Challenging question find if there is an element repeating himself n/k times

    - by gleb-pendler
    here how it's goes: You have an array size n and a constant k (whatever) you can assume the the array of int type tho it kind be of whatever type but just for the clearane let assume it's an integer. Describe an algorithm that finds if there is an element/s that repeat itself at least n/k times... if there is return one - do it in linear time running O(n) Imortent: now the catch do this algorithm or even pseuo-code using a constant usage of memory and running over the array only TWICE!!!

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