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• permutation of array

  - by davit-datuashvili

  for example i have array int a[]=new int[]{3,4,6,2,1}; I need list of all permutation such that if one is like this, {3,2,1,4,6}, others must not be the same i know that if length of array=n then there is n! possible combination please help me to write this algortihm what to do?

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• combination and permutation in C++

  - by skydoor

  What's the most widely used existing library in C++ to give all the combination and permutation of k elements out of n elements? I am not asking the algorithm but the existing library or methods. Thanks.

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• Permuting a binary tree without the use of lists

  - by Banang

  I need to find an algorithm for generating every possible permutation of a binary tree, and need to do so without using lists (this is because the tree itself carries semantics and restraints that cannot be translated into lists). I've found an algorithm that works for trees with the height of three or less, but whenever I get to greater hights, I loose one set of possible permutations per height added. Each node carries information about its original state, so that one node can determine if all possible permutations have been tried for that node. Also, the node carries information on weather or not it's been 'swapped', i.e. if it has seen all possible permutations of it's subtree. The tree is left-centered, meaning that the right node should always (except in some cases that I don't need to cover for this algorithm) be a leaf node, while the left node is always either a leaf or a branch. The algorithm I'm using at the moment can be described sort of like this: if the left child node has been swapped swap my right node with the left child nodes right node set the left child node as 'unswapped' if the current node is back to its original state swap my right node with the lowest left nodes' right node swap the lowest left nodes two childnodes set my left node as 'unswapped' set my left chilnode to use this as it's original state set this node as swapped return null return this; else if the left child has not been swapped if the result of trying to permute left child is null return the permutation of this node else return the permutation of the left child node if this node has a left node and a right node that are both leaves swap them set this node to be 'swapped' The desired behaviour of the algoritm would be something like this: branch / | branch 3 / | branch 2 / | 0 1 branch / | branch 3 / | branch 2 / | 1 0 <-- first swap branch / | branch 3 / | branch 1 <-- second swap / | 2 0 branch / | branch 3 / | branch 1 / | 0 2 <-- third swap branch / | branch 3 / | branch 0 <-- fourth swap / | 1 2 and so on... Sorry for the ridiculisly long and waddly explanation, would really, really apreciate any sort of help you guys could offer me. Thanks a bunch!

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• Sorting a vector of (double precision) reals and obtain their order

  - by Philipp

  Hello everyone, in C++ would like to sort a lengthy (2^20) vector of reals, obviously sort() does the trick. Having used R before I was used to the nice order() function which yields the permutation that leads to the sorted vector. Probably someone has done this in C++, maybe it's just my weak google-Fu that prevents me from finding it. And yeah, obivously my C++ newbness could stop me from spotting something straightforward. Example: x = {24, 55, 22, 1} then the permutation perm = {3, 2, 0, 1} maps the original x to the sorted x in ascending order. I can probably implement some bubble sort which does not only sort x but performs the same transpositions on the vector {0,1,2,...} and outputs both, but I believe someone must have thought about it and especially have done it efficiently. Thank you very much, Philipp

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• Python Permutation Program Flow help

  - by dsaccount1

  Hello world, i found this code at activestate, it takes a string and prints permutations of the string. I understand that its a recursive function but i dont really understand how it works, it'd be great if someone could walk me through the program flow, thanks a bunch! <pre><code> import sys def printList(alist, blist=[]): if not len(alist): print ''.join(blist) for i in range(len(alist)): blist.append(alist.pop(i)) printList(alist, blist) alist.insert(i, blist.pop()) if name == 'main': k='love' if len(sys.argv)1: k = sys.argv[1] printList(list(k))

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• Is it possible to enumerate for all permutations of two IEnumerables using linq

  - by James Hay

  I could do this using loops, but is there a way to take two IEnumerables, enumerate through all possible permutations and select an object that contains the permutation? I feel like this 'should' be possible but am not really sure what operators to use. Thanks James

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• 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.

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• Permutation on Rails Routes

  - by Kevin Sylvestre

  I currently have an application that for a set of parameters (location, category, budget, ...) a user can enter a "pretty" URL like: /location/canada/ontario /category/primary /budget/small Resulting in the respective parameters: params[:country] == 'canada' and params[:region] == 'ontario' params[:category] == 'primary' params[:budget] == 'small' I want to allow users to perform searches on multiple parameters at once (with each parameter optional). For example: /location/canada/ontario/category/primary/budget/small I understand that this can be achieved using URL parameters, but for SEO reasons I prefer to add the "pretty" parameters. Is this at all possible without listing all possible combination of routes (I have a large number of search-able fields)? I understand that route "globbing" maybe play a roll, but I am not sure how. Thanks.

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• in-place permutation of a array follows this rule

  - by Mgccl

  Suppose there is an array, we want to find everything in the odd index, and move it to the end. Everything in the even index move it to the beginning. The relative order of all odd index items and all even index items are preserved. Suppose the values of the array, a[i] = i, n is even. Then we have. 0,1,2,3,4,5,...,n-1 after the operation 0,2,4,6,...,n-2,1,3,5,7,...,n-1 Can this be done in-place and in O(n) time?

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• How to save an order (permutation) in an sql db

  - by Bendlas

  I have a tree structure in an sql table like so: CREATE TABLE containers ( container_id serial NOT NULL PRIMARY KEY, parent integer REFERENCES containers (container_id)) Now i want to define an ordering between nodes with the same parent. I Have thought of adding a node_index column, to ORDER BY, but that seem suboptimal, since that involves modifying the index of a lot of nodes when modifying the stucture. That could include adding, removing, reordering or moving nodes from some subtree to another. Is there a sql datatype for an ordered sequence, or an efficient way to emulate one? Doesn't need to be fully standard sql, I just need a solution for mssql and hopefully postgresql EDIT To make it clear, the ordering is arbitrary. Actually, the user will be able to drag'n'drop tree nodes in the GUI

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• 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++.

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• 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.

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• 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.

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• 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.

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• C++: recursively computer all permutaions of digits 0 - 9

  - by Nate

  I have a homework assignment where part of the requirement is to recursively compute all the permutations of integers 0 - 9. The professor actually gave us the algorithm for this part of the question. I've finished the rest of the assignment, but I can't get the permute function working...I'm implementing it exactly like it was shown on the assignment information. However, when I run it each permutation is repeated multiple times (and I'm not sure if I'm even getting all the correct permutations.) I think he must've made a mistake on the assignment instructions. I've been working on this for a couple of hours and can't seem to figure out where I'm going wrong. Can anybody help point me in the right direction? Here's the current code: void permute(int v[], int curr) { for (int i = curr; i < MAX; i++) { swap(v[i], v[curr]); permute(v, curr + 1); swap(v[curr], v[i]); } } EDIT: Actually, right after posting this I realized it has to do with the swap, right? Because right now i and curr are the same, so I'm swapping identical numbers. Hm, should it be swap(v[i], v[curr+1])?

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• 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

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• 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.

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• Permutatation algorithm without recursion? Java

  - by Andreas Hornig

  Hi, I would like to get all combination of a number without any repetation. Like 0.1.2, 0.2.1, 1.2.0, 1.0.2, 2.0.1, 2.1.0. I tried to find an easy scheme but couldn't find so I drawed a graph/tree for it and this screams to use recursion. But I would like to do it without, if this is possible. So could anyone please help mw how to do that? Thank you in advance, Andreas

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• 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

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• 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!)

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• Multiple range product in Python

  - by Tyr

  Is there a better way to do this: perms = product(range(1,7),range(1,7),range(1,7)) so that I can choose how many ranges I use? I want it to be equivalent to this, but scalable. def dice(num) if num == 1: perms = ((i,) for i in range(1,7)) elif num == 2: perms = product(range(1,7),range(1,7)) elif num == 3: perms = product(range(1,7),range(1,7),range(1,7)) #... and so on but I know there has to be a better way. I'm using it for counting dice outcomes. The actual code def dice(selection= lambda d: d[2]): perms = itertools.product(range(1,7),range(1,7),range(1,7)) return collections.Counter(((selection(sorted(i)) for i in perms))) where I can call it with a variety of selectors, like sum(d[0:2]) for the sum of the lowest 2 dice or d[1] to get the middle dice.

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• How to generate all variations with repetitions of a string?

  - by Svenstaro

  I want to generate all variations with repetitions of a string in C++ and I'd highly prefer a non-recursive algorithm. I've come up with a recursive algorithm in the past but due to the complexity (r^n) I'd like to see an iterative approach. I'm quite surprised that I wasn't able to find a solution to this problem anywhere on the web or on StackOverflow. I've come up with a Python script that does what I want as well: import itertools variations = itertools.product('ab', repeat=4) for variations in variations: variation_string = "" for letter in variations: variation_string += letter print variation_string Output: aaaa aaab aaba aabb abaa abab abba abbb baaa baab baba babb bbaa bbab bbba bbbb Ideally I'd like a C++ program that can produce the exact output, taking the exact same parameters. This is for learning purposes, it isn't homework. I wish my homework was like that.

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• 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!

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• Generate 10-digit number using a phone keypad

  - by srikfreak

  Given a phone keypad as shown below: 1 2 3 4 5 6 7 8 9 0 How many different 10-digit numbers can be formed starting from 1? The constraint is that the movement from 1 digit to the next is similar to the movement of the Knight in a chess game. For eg. if we are at 1 then the next digit can be either 6 or 8 if we are at 6 then the next digit can be 1, 7 or 0. Repetition of digits are allowed - 1616161616 is a valid number. Is there a polynomial time algorithm which solves this problem? The problem requires us to just give the count of 10-digit numbers and not necessarily list the numbers.

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• The algorithm used to generate recommendations in Google News?

  - by Siddhant

  Hi everyone. I'm study recommendation engines, and I went through the paper that defines how Google News generates recommendations to users for news items which might be of their interest, based on collaborative filtering. One interesting technique that they mention is Minhashing. I went through what it does, but I'm pretty sure that what I have is a fuzzy idea and there is a strong chance that I'm wrong. The following is what I could make out of it :- Collect a set of all news items. Define a hash function for a user. This hash function returns the index of the first item from the news items which this user viewed, in the list of all news items. Collect, say "n" number of such values, and represent a user with this list of values. Based on the similarity count between these lists, we can calculate the similarity between users as the number of common items. This reduces the number of comparisons a lot. Based on these similarity measures, group users into different clusters. This is just what I think it might be. In Step 2, instead of defining a constant hash function, it might be possible that we vary the hash function in a way that it returns the index of a different element. So one hash function could return the index of the first element from the user's list, another hash function could return the index of the second element from the user's list, and so on. So the nature of the hash function satisfying the minwise independent permutations condition, this does sound like a possible approach. Could anyone please confirm if what I think is correct? Or the minhashing portion of Google News Recommendations, functions in some other way? I'm new to internal implementations of recommendations. Any help is appreciated a lot. Thanks!

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