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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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  • Muliple Foreground Colors in Powershell in One Command.

    - by Mark Tomlin
    I want to output many different foreground colors with one statement. PS C:\> Write-Host "Red" -ForegroundColor Red Red This output is red. PS C:\> Write-Host "Blue" -ForegroundColor Blue Blue This output is blue. PS C:\> Write-Host "Red", "Blue" -ForegroundColor Red, Blue Red Blue This output is magenta, but I want the color to be Red for the word red, and blue for the word blue via the one command. How can I do that?

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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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  • O(log N) == O(1) - Why not?

    - by phoku
    Whenever I consider algorithms/data structures I tend to replace the log(N) parts by constants. Oh, I know log(N) diverges - but does it matter in real world applications? log(infinity) < 100 for all practical purposes. I am really curious for real world examples where this doesn't hold. To clarify: I understand O(f(N)) I am curious about real world examples where the asymptotic behaviour matters more than the constants of the actual performance. If log(N) can be replaced by a constant it still can be replaced by a constant in O( N log N). This question is for the sake of (a) entertainment and (b) to gather arguments to use if I run (again) into a controversy about the performance of a design.

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  • Time complexity O() of isPalindrome()

    - by Aran
    I have this method, isPalindrome(), and I am trying to find the time complexity of it, and also rewrite the code more efficiently. boolean isPalindrome(String s) { boolean bP = true; for(int i=0; i<s.length(); i++) { if(s.charAt(i) != s.charAt(s.length()-i-1)) { bP = false; } } return bP; } Now I know this code checks the string's characters to see whether it is the same as the one before it and if it is then it doesn't change bP. And I think I know that the operations are s.length(), s.charAt(i) and s.charAt(s.length()-i-!)). Making the time-complexity O(N + 3), I think? This correct, if not what is it and how is that figured out. Also to make this more efficient, would it be good to store the character in temporary strings?

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  • Linear time and quadratic time

    - by jasonline
    I'm just not sure... If you have a code that can be executed in either of the following complexities: (1) A sequence of O(n), like for example: two O(n) in sequence (2) 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? If no, 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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  • The limits of parallelism

    - by psihodelia
    Is it possible to solve a problem of O(n!) complexity within a reasonable time given infinite number of processing units and infinite space? The typical example of O(n!) problem is brute-force search: trying all permutations (ordered combinations).

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