Search Results

Search found 199 results on 8 pages for 'permutations'.

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

Page 1/8 | 1 2 3 4 5 6 7 8  | Next Page >

  • perl: generating permutations from a regular expression

    - by wibble
    I know you can generate all permutations from a list, using glob or Algorithm::Permute for example - but how do you generate all possible permutations from a regular expression? i want to do like: @perms = permute( "/\s[A-Z][0-9][0-9]/" ); sub permute( $regex ) { # code - put all permutations of above regex in a list return @list; }

    Read the article

  • All possible permutations of a set of lists in Python

    - by Ian Davis
    In Python I have a list of n lists, each with a variable number of elements. How can I create a single list containing all the possible permutations: For example [ [ a, b, c], [d], [e, f] ] I want [ [a, d, e] , [a, d, f], [b, d, e], [b, d, f], [c, d, e], [c, d, f] ] Note I don't know n in advance. I thought itertools.product would be the right approach but it requires me to know the number of arguments in advance

    Read the article

  • Maths Question: number of different permutations

    - by KingCong
    This is more of a maths question than programming but I figure a lot of people here are pretty good at maths! :) My question is: Given a 9 x 9 grid (81 cells) that must contain the numbers 1 to 9 each exactly 9 times, how many different grids can be produced. The order of the numbers doesn't matter, for example the first row could contain nine 1's etc. This is related to Sudoku and we know the number of valid Sudoku grids is 6.67×10^21, so since my problem isn't constrained like Sudoku by having to have each of the 9 numbers in each row, column and box then the answer should be greater than 6.67×10^21. My first thought was that the answer is 81! however on further reflection this assume that the 81 number possible for each cell are different, distinct number. They are not, there are 81 possible numbers for each cell but only 9 possible different numbers. My next thought was then that each of the cells in the first row can be any number between 1 and 9. If by chance the first row happened to be all the same number, say all 1s, then each cell in the second row could only have 8 possibilites, 2-9. If this continued down until the last row then number of different permutations could be calculated by 9^2 * 8^2 * 7^2 ..... * 1^2. However this doesn't work if each row doesn't contain 9 of the same number. It's been quite a while since I studied this stuff and I can't think of a way to work it out, I'd appreciate any help anyone can offer.

    Read the article

  • Trying all permutations

    - by Malfist
    For my program, I am trying to assist the user, and reduce his or her work load. There are four input numbers. There is also an indeterminate amount of numbers they can be applied too. For example, they four input numbers could be {4,7,3,2} and the numbers they can be applied to are {4,9,23} The result should be: 4 (input) was applied to 4, leaving the sets looking like: {0,7,3,2} and then 7,2 (input) are applied to 9 leaving the sets looking like: {0,0,3,0} and {0,0,23} and because 3 or any other permutation including 3 does not match 23, 3 remains. How would I do this?

    Read the article

  • Generate Permutations of a List

    - by Eric Mercer
    I'm writing a function that takes a list and returns a list of permutations of the argument. I know how to do it by using a function that removes an element and then recursively use that function to generate all permutations. I now have a problem where I want to use the following function: (define (insert-everywhere item lst) (define (helper item L1 L2) (if (null? L2) (cons (append L1 (cons item '())) '()) (cons (append L1 (cons item L2)) (helper item (append L1 (cons (car L2) '())) (cdr L2))))) (helper item '() lst)) This function will insert the item into every possible location of the list, like the following: (insert-everywhere 1 '(a b)) will get: '((1 a b) (a 1 b) (a b 1)) How would I use this function to get all permutations of a list? I now have: (define (permutations lst) (if (null? lst) '() (insert-helper (car lst) (permutations (cdr lst))))) (define (insert-helper item lst) (cond ((null? lst) '()) (else (append (insert-everywhere item (car lst)) (insert-helper item (cdr lst)))))) but doing (permutations '(1 2 3)) just returns the empty list '().

    Read the article

  • What are the Ruby equivalent of Python itertools, esp. combinations/permutations/groupby?

    - by Amadeus
    Python's itertools module provides a lots of goodies with respect to processing an iterable/iterator by use of generators. For example, permutations(range(3)) --> 012 021 102 120 201 210 combinations('ABCD', 2) --> AB AC AD BC BD CD [list(g) for k, g in groupby('AAAABBBCCD')] --> AAAA BBB CC D What are the equivalent in Ruby? By equivalent, I mean fast and memory efficient (Python's itertools module is written in C).

    Read the article

  • Given an array of integers [x0 x1 x2], how do you calculate all possible permutations from [0 0 0] t

    - by user319951
    I am writing a program that takes in an ArrayList and I need to calculate all possible permutations starting with a list of zeroes, up to the value in the corresponding input list. Does anyone know how to iteratively calculate these values? For example, given [ 1 2 ] as input, it should find and store the following lists: [0 0], [1 0], [1 1], [1 2], [0 1], [0 2] Thanks!

    Read the article

  • algorithm for python itertools.permutations

    - by zaharpopov
    Can someone please explain algorithm for itertools.permutations routine in Python standard lib 2.6? I see its code in the documentation but don't undestand why it work? Thanks Code is: def permutations(iterable, r=None): # permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC # permutations(range(3)) --> 012 021 102 120 201 210 pool = tuple(iterable) n = len(pool) r = n if r is None else r if r > n: return indices = range(n) cycles = range(n, n-r, -1) yield tuple(pool[i] for i in indices[:r]) while n: for i in reversed(range(r)): cycles[i] -= 1 if cycles[i] == 0: indices[i:] = indices[i+1:] + indices[i:i+1] cycles[i] = n - i else: j = cycles[i] indices[i], indices[-j] = indices[-j], indices[i] yield tuple(pool[i] for i in indices[:r]) break else: return

    Read the article

  • Permutations of Varying Size

    - by waiwai933
    I'm trying to write a function in PHP that gets all permutations of all possible sizes. I think an example would be the best way to start off: $my_array = array(1,1,2,3); Possible permutations of varying size: 1 1 // * See Note 2 3 1,1 1,2 1,3 // And so forth, for all the sets of size 2 1,1,2 1,1,3 1,2,1 // And so forth, for all the sets of size 3 1,1,2,3 1,1,3,2 // And so forth, for all the sets of size 4 Note: I don't care if there's a duplicate or not. For the purposes of this example, all future duplicates have been omitted. What I have so far in PHP: function getPermutations($my_array){ $permutation_length = 1; $keep_going = true; while($keep_going){ while($there_are_still_permutations_with_this_length){ // Generate the next permutation and return it into an array // Of course, the actual important part of the code is what I'm having trouble with. } $permutation_length++; if($permutation_length>count($my_array)){ $keep_going = false; } else{ $keep_going = true; } } return $return_array; } The closest thing I can think of is shuffling the array, picking the first n elements, seeing if it's already in the results array, and if it's not, add it in, and then stop when there are mathematically no more possible permutations for that length. But it's ugly and resource-inefficient. Any pseudocode algorithms would be greatly appreciated. Also, for super-duper (worthless) bonus points, is there a way to get just 1 permutation with the function but make it so that it doesn't have to recalculate all previous permutations to get the next? For example, I pass it a parameter 3, which means it's already done 3 permutations, and it just generates number 4 without redoing the previous 3? (Passing it the parameter is not necessary, it could keep track in a global or static). The reason I ask this is because as the array grows, so does the number of possible combinations. Suffice it to say that one small data set with only a dozen elements grows quickly into the trillions of possible combinations and I don't want to task PHP with holding trillions of permutations in its memory at once.

    Read the article

  • Permutations distinct under given symmetry (Mathematica 8 group theory)

    - by Yaroslav Bulatov
    Given a list of integers like {2,1,1,0} I'd like to list all permutations of that list that are not equivalent under given group. For instance, using symmetry of the square, the result would be {{2, 1, 1, 0}, {2, 1, 0, 1}}. Approach below (Mathematica 8) generates all permutations, then weeds out the equivalent ones. I can't use it because I can't afford to generate all permutations, is there a more efficient way? Update: actually, the bottleneck is in DeleteCases. The following list {2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 0, 0, 0} has about a million permutations and takes 0.1 seconds to compute. Apparently there are supposed to be 1292 orderings after removing symmetries, but my approach doesn't finish in 10 minutes removeEquivalent[{}] := {}; removeEquivalent[list_] := ( Sow[First[list]]; equivalents = Permute[First[list], #] & /@ GroupElements[group]; DeleteCases[list, Alternatives @@ equivalents] ); nonequivalentPermutations[list_] := ( reaped = Reap@FixedPoint[removeEquivalent, Permutations@list]; reaped[[2, 1]] ); group = DihedralGroup[4]; nonequivalentPermutations[{2, 1, 1, 0}]

    Read the article

  • What are some practical uses of generating all permutations of a list, such as ['a', 'b', 'c'] ?

    - by Jian Lin
    I was asked by somebody in an interview for web front end job, to write a function that generates all permutation of a string, such as "abc" (or consider it ['a', 'b', 'c']). so the expected result from the function, when given ['a', 'b', 'c'], is abc acb bac bca cab cba Actually in my past 20 years of career, I have never needed to do something like that, especially when doing front end work for web programming. What are some practical use of this problem nowadays, in web programming, front end or back end, I wonder? As a side note, I kind of feel that expecting a result in 3 minutes might be "either he gets it or he doesn't", especially I was thinking of doing it by a procedural, non-recursive way at first. After the interview, I spent another 10 minutes and thought of how to do it using recursion, but expecting it to be solved within 3 minutes... may not be a good test of how qualified he is, especially for front end work.

    Read the article

  • All Permutations of a string when corresponding characters are not in the same place

    - by r20rock
    I need all possible permutations of a given string such that no character should remain at the same place as in the input string. Eg. : for input "ask" Output: all possible permutaions like "ksa", "kas"... such that 'a' is not in the 1st position , 's' is not in the 2nd positions and so on... in any permutation. I only need the count of such possible permutations I can do this by generating all permutations and filtering them but I need a very efficient way of doing this. All characters in the string are "UNIQUE" Preferred language C++.

    Read the article

  • Combinations and Permutations in F#

    - by Noldorin
    I've recently written the following combinations and permutations functions for an F# project, but I'm quite aware they're far from optimised. /// Rotates a list by one place forward. let rotate lst = List.tail lst @ [List.head lst] /// Gets all rotations of a list. let getRotations lst = let rec getAll lst i = if i = 0 then [] else lst :: (getAll (rotate lst) (i - 1)) getAll lst (List.length lst) /// Gets all permutations (without repetition) of specified length from a list. let rec getPerms n lst = match n, lst with | 0, _ -> seq [[]] | _, [] -> seq [] | k, _ -> lst |> getRotations |> Seq.collect (fun r -> Seq.map ((@) [List.head r]) (getPerms (k - 1) (List.tail r))) /// Gets all permutations (with repetition) of specified length from a list. let rec getPermsWithRep n lst = match n, lst with | 0, _ -> seq [[]] | _, [] -> seq [] | k, _ -> lst |> Seq.collect (fun x -> Seq.map ((@) [x]) (getPermsWithRep (k - 1) lst)) // equivalent: | k, _ -> lst |> getRotations |> Seq.collect (fun r -> List.map ((@) [List.head r]) (getPermsWithRep (k - 1) r)) /// Gets all combinations (without repetition) of specified length from a list. let rec getCombs n lst = match n, lst with | 0, _ -> seq [[]] | _, [] -> seq [] | k, (x :: xs) -> Seq.append (Seq.map ((@) [x]) (getCombs (k - 1) xs)) (getCombs k xs) /// Gets all combinations (with repetition) of specified length from a list. let rec getCombsWithRep n lst = match n, lst with | 0, _ -> seq [[]] | _, [] -> seq [] | k, (x :: xs) -> Seq.append (Seq.map ((@) [x]) (getCombsWithRep (k - 1) lst)) (getCombsWithRep k xs) Does anyone have any suggestions for how these functions (algorithms) can be sped up? I'm particularly interested in how the permutation (with and without repetition) ones can be improved. The business involving rotations of lists doesn't look too efficient to me in retrospect. Update Here's my new implementation for the getPerms function, inspired by Tomas's answer. Unfortunately, it's not really any fast than the existing one. Suggestions? let getPerms n lst = let rec getPermsImpl acc n lst = seq { match n, lst with | k, x :: xs -> if k > 0 then for r in getRotations lst do yield! getPermsImpl (List.head r :: acc) (k - 1) (List.tail r) if k >= 0 then yield! getPermsImpl acc k [] | 0, [] -> yield acc | _, [] -> () } getPermsImpl List.empty n lst

    Read the article

  • Permutations with extra restrictions

    - by Full Decent
    I have a set of items, for example: {1,1,1,2,2,3,3,3}, and a restricting set of sets, for example {{3},{1,2},{1,2,3},{1,2,3},{1,2,3},{1,2,3},{2,3},{2,3}. I am looking for permutations of items, but the first element must be 3, and the second must be 1 or 2, etc. One such permutation that fits is: {3,1,1,1,2,2,3} Is there an algorithm to count all permutations for this problem in general? Is there a name for this type of problem? For illustration, I know how to solve this problem for certain types of "restricting sets". Set of items: {1,1,2,2,3}, Restrictions {{1,2},{1,2,3},{1,2,3},{1,2},{1,2}}. This is equal to 2!/(2-1)!/1! * 4!/2!/2!. Effectively permuting the 3 first, since it is the most restrictive and then permuting the remaining items where there is room.

    Read the article

  • permutations gone wrong

    - by vbNewbie
    I have written code to implement an algorithm I found on string permutations. What I have is an arraylist of words ( up to 200) and I need to permutate the list in levels of 5. Basically group the string words in fives and permutated them. What I have takes the first 5 words generates the permutations and ignores the rest of the arraylist? Any ideas appreciated. Private Function permute(ByVal chunks As ArrayList, ByVal k As Long) As ArrayList ReDim ItemUsed(k) pno = 0 Permutate(k, 1) Return chunks End Function Private Shared Sub Permutate(ByVal K As Long, ByVal pLevel As Long) Dim i As Long, Perm As String Perm = pString ' Save the current Perm ' for each value currently available For i = 1 To K If Not ItemUsed(i) Then If pLevel = 1 Then pString = chunks.Item(i) 'pString = inChars(i) Else pString = pString & chunks.Item(i) 'pString += inChars(i) End If If pLevel = K Then 'got next Perm pno = pno + 1 SyncLock outfile outfile.WriteLine(pno & " = " & pString & vbCrLf) End SyncLock outfile.Flush() Exit Sub End If ' Mark this item unavailable ItemUsed(i) = True ' gen all Perms at next level Permutate(K, pLevel + 1) ' Mark this item free again ItemUsed(i) = False ' Restore the current Perm pString = Perm End If Next K above is = to 5 for the number of words in one permutation but when I change the for loop to the arraylist size I get an error of index out of bounds

    Read the article

  • Generating All Permutations of Character Combinations when # of arrays and length of each array are

    - by Jay
    Hi everyone, I'm not sure how to ask my question in a succinct way, so I'll start with examples and expand from there. I am working with VBA, but I think this problem is non language specific and would only require a bright mind that can provide a pseudo code framework. Thanks in advance for the help! Example: I have 3 Character Arrays Like So: Arr_1 = [X,Y,Z] Arr_2 = [A,B] Arr_3 = [1,2,3,4] I would like to generate ALL possible permutations of the character arrays like so: XA1 XA2 XA3 XA4 XB1 XB2 XB3 XB4 YA1 YA2 . . . ZB3 ZB4 This can be easily solved using 3 while loops or for loops. My question is how do I solve for this if the # of arrays is unknown and the length of each array is unknown? So as an example with 4 character arrays: Arr_1 = [X,Y,Z] Arr_2 = [A,B] Arr_3 = [1,2,3,4] Arr_4 = [a,b] I would need to generate: XA1a XA1b XA2a XA2b XA3a XA3b XA4a XA4b . . . ZB4a ZB4b So the Generalized Example would be: Arr_1 = [...] Arr_2 = [...] Arr_3 = [...] . . . Arr_x = [...] Is there a way to structure a function that will generate an unknown number of loops and loop through the length of each array to generate the permutations? Or maybe there's a better way to think about the problem? Thanks Everyone!

    Read the article

  • Calculating permutations in F#

    - by Benjol
    Inspired by this question and answer, how do I create a generic permutations algorithm in F#? Google doesn't give any useful answers to this. EDIT: I provide my best answer below, but I suspect that Tomas's is better (certainly shorter!)

    Read the article

  • Find valid assignments of integers in arrays (permutations with given order)

    - by evident
    Hi everybody! I am having a general problem finding a good algorithm for generating each possible assignment for some integers in different arrays. Lets say I have n arrays and m numbers (I can have more arrays than numbers, more numbers than arrays or as much arrays as numbers). As an example I have the numbers 1,2,3 and three arrays: { }, { }, { } Now I would like to find each of these solutions: {1,2,3}, { }, { } { }, {1,2,3}, { } { }, { }, {1,2,3} {1,2}, {3}, { } {1,2}, { }, {3} { }, {1,2}, {3} {1}, {2,3}, { } {1}, { }, {2,3} { }, {1}, {2,3} {1}, {2}, {3} So basically I would like to find each possible combination to assign the numbers to the different arrays with keeping the order. So as in the example the 1 always needs to come before the others and so on... I want to write an algorithm in C++/Qt to find all these valid combinations. Does anybody have an approach for me on how to handle this problem? How would I generate these permutations?

    Read the article

  • list permutations in haskell

    - by turingcomplete
    So I'm new to haskell and I've been playing with it for a while now. I want to get my function that outputs all list permutations to work. I have written 2 implementations, one works well, the other is giving me an error. Any help would be awesome. This is the first (working) implementation: permute [] = [[]] permute xs = [y| x <- xs, y <- map (x:) $ permute $ delete x xs] This one is giving me an error: permute [] = [[]] permute xs = map (\x -> map (x:) $ permute $ delete x xs) xs and here's the error message: Occurs check: cannot construct the infinite type: t0 = [t0] Expected type: [t0] Actual type: [[t0]] In the expression: map (x :) $ permute $ delete x xs In the first argument of `map', namely `(\ x -> map (x :) $ permute $ delete x xs)' I'd appreciate if someone could explain why I'm getting this error. Thanks

    Read the article

  • Count in base 2, 3, 4 etc in Java and output all permutations

    - by tree-hacker
    I want to write a function in Java that takes as input an integer and outputs every possible permutation of numbers up to that integer. For example: f(1) 0 f(2) should output: 00 01 10 11 f(3) should output: 000 001 002 010 011 012 020 021 022 100 .... 220 221 222 That is it should output all 27 permutations of the digits of the numbers 0, 1, 2. f(4) should output 0000 0001 0002 0003 0010 ... 3330 3331 3332 3333 f(4) should output 00000 00001 ... 44443 44444 I have been trying to solve this problem but cannot seem to work out how to do it and keep getting confused by how many loops I need. Does anyone know how to solve this problem? Thanks in advance.

    Read the article

  • Generating the permutations from a number of Characters

    - by adam08
    I'm working on a predictive text solution and have all the words being retrieved from a Trie based on input for a certain string of characters, i.e. "at" will give all words formed with "at" as a prefix. The problem that I have now, is that we are also supposed to return all other possibilities from pressing these 2 buttons, Button 2 and button 8 on the mobile phone, which would also give words formed with, "au, av, bt, bu, bv, ct, cu, cv" (most of which won't have any actual words. Can anyone suggest a solution and how I would go about doing this for calculating the different permutations? (at the moment, I'm prompting the user to enter the prefix (not using a GUI right now)

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

1 2 3 4 5 6 7 8  | Next Page >