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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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  • ASM programming, how to use loop?

    - by chris
    Hello. Im first time here.I am a college student. I've created a simple program by using assembly language. And im wondering if i can use loop method to run it almost samething as what it does below the program i posted. and im also eager to find someome who i can talk through MSN messanger so i can ask you questions right away.(if possible) ok thank you .MODEL small .STACK 400h .data prompt db 10,13,'Please enter a 3 digit number, example 100:',10,13,'$' ;10,13 cause to go to next line first_digit db 0d second_digit db 0d third_digit db 0d Not_prime db 10,13,'This number is not prime!',10,13,'$' prime db 10,13,'This number is prime!',10,13,'$' question db 10,13,'Do you want to contine Y/N $' counter dw 0d number dw 0d half dw ? .code Start: mov ax, @data ;establish access to the data segment mov ds, ax mov number, 0d LetsRoll: mov dx, offset prompt ; print the string (please enter a 3 digit...) mov ah, 9h int 21h ;execute ;read FIRST DIGIT mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov first_digit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, doubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer mov cx, 100d ;This is so we can calculate 100*1st digit +10*2nd digit + 3rd digit mul cx ;start to accumulate the 3 digit number in the variable imul cx ;it is understood that the other operand is ax ;AND that the result will use both dx::ax ;but we understand that dx will contain only leading zeros add number, ax ;save ;variable <number> now contains 1st digit * 10 ;---------------------------------------------------------------------- ;read SECOND DIGIT, multiply by 10 and add in mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov second_digit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, boubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer mov cx, 10d ;continue to accumulate the 3 digit number in the variable mul cx ;it is understood that the other operand is ax, containing first digit ;AND that the result will use both dx::ax ;but we understand that dx will contain only leading zeros. Ignore them add number, ax ;save -- nearly finished ;variable <number> now contains 1st digit * 100 + second digit * 10 ;---------------------------------------------------------------------- ;read THIRD DIGIT, add it in (no multiplication this time) mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov third_digit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, boubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer add number, ax ;Both my variable number and ax are 16 bits, so equal size mov ax, number ;copy contents of number to ax mov cx, 2h div cx ;Divide by cx mov half, ax ;copy the contents of ax to half mov cx, 2h; mov ax, number; ;copy numbers to ax xor dx, dx ;flush dx jmp prime_check ;jump to prime check print_question: mov dx, offset question ;print string (do you want to continue Y/N?) mov ah, 9h int 21h ;execute mov ah, 1h int 21h ;execute cmp al, 4eh ;compare je Exit ;jump to exit cmp al, 6eh ;compare je Exit ;jump to exit cmp al, 59h ;compare je Start ;jump to start cmp al, 79h ;compare je Start ;jump to start prime_check: div cx; ;Divide by cx cmp dx, 0h ;reset the value of dx je print_not_prime ;jump to not prime xor dx, dx; ;flush dx mov ax, number ;copy the contents of number to ax cmp cx, half ;compare half with cx je print_prime ;jump to print prime section inc cx; ;increment cx by one jmp prime_check ;repeat the prime check print_prime: mov dx, offset prime ;print string (this number is prime!) mov ah, 9h int 21h ;execute jmp print_question ;jumps to question (do you want to continue Y/N?) this is for repeat print_not_prime: mov dx, offset Not_prime ;print string (this number is not prime!) mov ah, 9h int 21h ;execute jmp print_question ;jumps to question (do you want to continue Y/N?) this is for repeat Exit: mov ah, 4ch int 21h ;execute exit END Start

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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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  • 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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  • 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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  • 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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  • 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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