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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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  • 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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  • Python: (sampling with replacement): efficient algorithm to extract the set of UNIQUE N-tuples from a set

    - by Homunculus Reticulli
    I have a set of items, from which I want to select DISSIMILAR tuples (more on the definition of dissimilar touples later). The set could contain potentially several thousand items, although typically, it would contain only a few hundreds. I am trying to write a generic algorithm that will allow me to select N items to form an N-tuple, from the original set. The new set of selected N-tuples should be DISSIMILAR. A N-tuple A is said to be DISSIMILAR to another N-tuple B if and only if: Every pair (2-tuple) that occurs in A DOES NOT appear in B Note: For this algorithm, A 2-tuple (pair) is considered SIMILAR/IDENTICAL if it contains the same elements, i.e. (x,y) is considered the same as (y,x). This is a (possible variation on the) classic Urn Problem. A trivial (pseudocode) implementation of this algorithm would be something along the lines of def fetch_unique_tuples(original_set, tuple_size): while True: # randomly select [tuple_size] items from the set to create first set # create a key or hash from the N elements and store in a set # store selected N-tuple in a container if end_condition_met: break I don't think this is the most efficient way of doing this - and though I am no algorithm theorist, I suspect that the time for this algorithm to run is NOT O(n) - in fact, its probably more likely to be O(n!). I am wondering if there is a more efficient way of implementing such an algo, and preferably, reducing the time to O(n). Actually, as Mark Byers pointed out there is a second variable m, which is the size of the number of elements being selected. This (i.e. m) will typically be between 2 and 5. Regarding examples, here would be a typical (albeit shortened) example: original_list = ['CAGG', 'CTTC', 'ACCT', 'TGCA', 'CCTG', 'CAAA', 'TGCC', 'ACTT', 'TAAT', 'CTTG', 'CGGC', 'GGCC', 'TCCT', 'ATCC', 'ACAG', 'TGAA', 'TTTG', 'ACAA', 'TGTC', 'TGGA', 'CTGC', 'GCTC', 'AGGA', 'TGCT', 'GCGC', 'GCGG', 'AAAG', 'GCTG', 'GCCG', 'ACCA', 'CTCC', 'CACG', 'CATA', 'GGGA', 'CGAG', 'CCCC', 'GGTG', 'AAGT', 'CCAC', 'AACA', 'AATA', 'CGAC', 'GGAA', 'TACC', 'AGTT', 'GTGG', 'CGCA', 'GGGG', 'GAGA', 'AGCC', 'ACCG', 'CCAT', 'AGAC', 'GGGT', 'CAGC', 'GATG', 'TTCG'] Select 3-tuples from the original list should produce a list (or set) similar to: [('CAGG', 'CTTC', 'ACCT') ('CAGG', 'TGCA', 'CCTG') ('CAGG', 'CAAA', 'TGCC') ('CAGG', 'ACTT', 'ACCT') ('CAGG', 'CTTG', 'CGGC') .... ('CTTC', 'TGCA', 'CAAA') ] [[Edit]] Actually, in constructing the example output, I have realized that the earlier definition I gave for UNIQUENESS was incorrect. I have updated my definition and have introduced a new metric of DISSIMILARITY instead, as a result of this finding.

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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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  • Given a number N, find the number of ways to write it as a sum of two or more consecutive integers

    - by hilal
    Here is the problem (Given a number N, find the number of ways to write it as a sum of two or more consecutive integers) and example 15 = 7+8, 1+2+3+4+5, 4+5+6 I solved with math like that : a + (a + 1) + (a + 2) + (a + 3) + ... + (a + k) = N (k + 1)*a + (1 + 2 + 3 + ... + k) = N (k + 1)a + k(k+1)/2 = N (k + 1)*(2*a + k)/2 = N Then check that if N divisible by (k+1) and (2*a+k) then I can find answer in O(N) time Here is my question how can you solve this by dynamic-programming ? and what is the complexity (O) ? P.S : excuse me, if it is a duplicate question. I searched but I can find

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  • Link failure with either abnormal memory consumption or LNK1106 in Visual Studio 2005.

    - by Corvin
    Hello, I am trying to build a solution for windows XP in Visual Studio 2005. This solution contains 81 projects (static libs, exe's, dlls) and is being successfully used by our partners. I copied the solution bundle from their repository and tried setting it up on 3 similar machines of people in our group. I was successful on two machines and the solution failed to build on my machine. The build on my machine encountered two problems: During a simple build creation of the biggest static library (about 522Mb in debug mode) would fail with the message "13libd\ui1d.lib : fatal error LNK1106: invalid file or disk full: cannot seek to 0x20101879" Full solution rebuild creates this library, however when it comes to linking the library to main .exe file, devenv.exe spawns link.exe which consumes about 80Mb of physical memory and 250MB of virtual and spawns another link.exe, which does the same. This goes on until the system runs out of memory. On PCs of my colleagues where successful build could be performed, there is only one link.exe process which uses all the memory required for linking (about 500Mb physical). There is a plenty of hard drive space on my machine and the file system is NTFS. All three of our systems are similar - Core2Quad processors, 4Gb of RAM, Windows XP SP3. We are using Visual studio installed from the same source. I tried using a different RAM and CPU, using dedicated graphics adapter to eliminate possibility of video memory sharing influencing the build, putting solution files to different location, using different versions of VS 2005 (Professional, Standard and Team Suite), changing the amount of available virtual memory, running memtest86 and building the project from scratch (i.e. a clean bundle). I have read what MSDN says about LNK1106, none of the cases apply to me except for maybe "out of heap space", however I am not sure how I should fight this. The only idea that I have left is reinstalling the OS, however I am not sure that it would help and I am not sure that my situation wouldn't repeat itself on a different machine. Would anyone have any sort of advice for me? Thanks

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  • Python: (sampling with replacement): efficient algorithm to extract the set of DISSIMILAR N-tuples from a set

    - by Homunculus Reticulli
    I have a set of items, from which I want to select DISSIMILAR tuples (more on the definition of dissimilar touples later). The set could contain potentially several thousand items, although typically, it would contain only a few hundreds. I am trying to write a generic algorithm that will allow me to select N items to form an N-tuple, from the original set. The new set of selected N-tuples should be DISSIMILAR. A N-tuple A is said to be DISSIMILAR to another N-tuple B if and only if: Every pair (2-tuple) that occurs in A DOES NOT appear in B Note: For this algorithm, A 2-tuple (pair) is considered SIMILAR/IDENTICAL if it contains the same elements, i.e. (x,y) is considered the same as (y,x). This is a (possible variation on the) classic Urn Problem. A trivial (pseudocode) implementation of this algorithm would be something along the lines of def fetch_unique_tuples(original_set, tuple_size): while True: # randomly select [tuple_size] items from the set to create first set # create a key or hash from the N elements and store in a set # store selected N-tuple in a container if end_condition_met: break I don't think this is the most efficient way of doing this - and though I am no algorithm theorist, I suspect that the time for this algorithm to run is NOT O(n) - in fact, its probably more likely to be O(n!). I am wondering if there is a more efficient way of implementing such an algo, and preferably, reducing the time to O(n). Actually, as Mark Byers pointed out there is a second variable m, which is the size of the number of elements being selected. This (i.e. m) will typically be between 2 and 5. Regarding examples, here would be a typical (albeit shortened) example: original_list = ['CAGG', 'CTTC', 'ACCT', 'TGCA', 'CCTG', 'CAAA', 'TGCC', 'ACTT', 'TAAT', 'CTTG', 'CGGC', 'GGCC', 'TCCT', 'ATCC', 'ACAG', 'TGAA', 'TTTG', 'ACAA', 'TGTC', 'TGGA', 'CTGC', 'GCTC', 'AGGA', 'TGCT', 'GCGC', 'GCGG', 'AAAG', 'GCTG', 'GCCG', 'ACCA', 'CTCC', 'CACG', 'CATA', 'GGGA', 'CGAG', 'CCCC', 'GGTG', 'AAGT', 'CCAC', 'AACA', 'AATA', 'CGAC', 'GGAA', 'TACC', 'AGTT', 'GTGG', 'CGCA', 'GGGG', 'GAGA', 'AGCC', 'ACCG', 'CCAT', 'AGAC', 'GGGT', 'CAGC', 'GATG', 'TTCG'] # Select 3-tuples from the original list should produce a list (or set) similar to: [('CAGG', 'CTTC', 'ACCT') ('CAGG', 'TGCA', 'CCTG') ('CAGG', 'CAAA', 'TGCC') ('CAGG', 'ACTT', 'ACCT') ('CAGG', 'CTTG', 'CGGC') .... ('CTTC', 'TGCA', 'CAAA') ] [[Edit]] Actually, in constructing the example output, I have realized that the earlier definition I gave for UNIQUENESS was incorrect. I have updated my definition and have introduced a new metric of DISSIMILARITY instead, as a result of this finding.

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  • Algorithm to determine if array contains n...n+m?

    - by Kyle Cronin
    I saw this question on Reddit, and there were no positive solutions presented, and I thought it would be a perfect question to ask here. This was in a thread about interview questions: Write a method that takes an int array of size m, and returns (True/False) if the array consists of the numbers n...n+m-1, all numbers in that range and only numbers in that range. The array is not guaranteed to be sorted. (For instance, {2,3,4} would return true. {1,3,1} would return false, {1,2,4} would return false. The problem I had with this one is that my interviewer kept asking me to optimize (faster O(n), less memory, etc), to the point where he claimed you could do it in one pass of the array using a constant amount of memory. Never figured that one out. Along with your solutions please indicate if they assume that the array contains unique items. Also indicate if your solution assumes the sequence starts at 1. (I've modified the question slightly to allow cases where it goes 2, 3, 4...) edit: I am now of the opinion that there does not exist a linear in time and constant in space algorithm that handles duplicates. Can anyone verify this? The duplicate problem boils down to testing to see if the array contains duplicates in O(n) time, O(1) space. If this can be done you can simply test first and if there are no duplicates run the algorithms posted. So can you test for dupes in O(n) time O(1) space?

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