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• How much does precomputation (matching a series of strings and their permutations with a set number

  - by nipun

  Consider a typical slots machine with n reels(say reel1: a,b,c,d,w1,d,b, ..etc). On play we generate a concatenated string of n objects (like for above, chars) We have a paytable which lists winning strings with payout amounts. The problem is a wild character (list of wilds: w1,w2) which can replace {w1:a,b,c},{w2:a} ..etc. Is it really worthwhile to have all possible winning strings permutations with the wilds precomputed and used or simply at the time of occurance, generate all combinations with the pattern in hand accordingly. I did'nt really see much difference initially, but now if I need to scale the machine to handle 11+ reels with a much higher concentration of wilds than previously, I need to figure out the exact approach for this particular bit. Any ideas will be really appreciated :)

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• Algorithm to generate all possible letter combinations of given string down to 2 letters

  - by Alan

  Algorithm to generate all possible letter combinations of given string down to 2 letters Trying to create an Anagram solver in AS3, such as this one found here: http://homepage.ntlworld.com/adam.bozon/anagramsolver.htm I'm having a problem wrapping my brain around generating all possible letter combinations for the various lengths of strings. If I was only generating permutations for a fixed length, it wouldn't be such a problem for me... but I'm looking to reduce the length of the string and obtain all the possible permutations from the original set of letters for a string with a max length smaller than the original string. For example, say I want a string length of 2, yet I have a 3 letter string of “abc”, the output would be: ab ac ba bc ca cb. Ideally the algorithm would produce a complete list of possible combinations starting with the original string length, down to the smallest string length of 2. I have a feeling there is probably a small recursive algorithm to do this, but can't wrap my brain around it. I'm working in AS3. Thanks!

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• Prolog: permutations check and multiple answers

  - by Adrian

  Continuing to learn prolog, I'm trying to write a permutation(L1, L2) predicate. It should return true, only if L1 can be made up of all elements in L2. My code so far is the following: permutation([], []). permutation([H|T], L2) :- remove(L2, H, R), permutation(T, R). Assuming that the predicate remove(L1, X, R) functions correctly, and it removes X from L1, I do not get correct results: ?- permutation([1],[1]). true ; false. ?- permutation([1, 2],[1]). true ; false. ?- permutation([1, 2],[1, 2]). true ; false. ?- permutation([1],[1, 2]). false. What am I missing. Subquestion: What happens when the remove predicate returns two answers? My implementation returns the new, correct list, and then after pressing ; it returns the original list. Which answer does permutation predicate use? ?- remove([1,2,3], 3, R). R = [1, 2] ; R = [1, 2, 3].

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• Counting the amount of letters in all permutations of words in R

  - by Rhodo

  I have some words: shapes<- c("Square", "Triangle","Octagon","Hexagon") I want to arrange them in pairs: shapescount<-combn(shapes, 2) shapescount [,1] [,2] [,3] [,4] [,5] [,6] [1,] "Square" "Square" "Square" "Triangle" "Triangle" "Octagon" [2,] "Triangle" "Octagon" "Hexagon" "Octagon" "Hexagon" "Hexagon" I want to count each of the groupings of the letters in the pairs, for instance first pair is "6" for "Square" and "8" for "Triangle" giving me "14" for the first pair, and so on.

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• Permutations in python 2.5.2

  - by flpgdt

  Hi, I have a list of numbers for input, e.g. 671.00 1,636.00 436.00 9,224.00 and I want to generate all possible sums with a way to id it for output, e.g.: 671.00 + 1,636.00 = 2,307.00 671.00 + 436.00 = 1,107.00 671.00 + 9,224.00 = 9,224.00 671.00 + 1,636.00 + 436.00 = 2,743.00 ... and I would like to do it in Python My current constrains are: a) I'm just learning python now (that's part of the idea) b) I will have to use Python 2.5.2 (no intertools) I think I have found a piece of code that may help: def all_perms(str): if len(str) <=1: yield str else: for perm in all_perms(str[1:]): for i in range(len(perm)+1): #nb str[0:1] works in both string and list contexts yield perm[:i] + str[0:1] + perm[i:] ( from these guys ) But I'm not sure how to use it in my propose. Could someone trow some tips and pieces of code of help? cheers, f.

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• Generate all permutations with sort constraint

  - by Moos Hueting

  Hi! I have a list consisting of other lists and some zeroes, for example: x = [[1, 1, 2], [1, 1, 1, 2], [1, 1, 2], 0, 0, 0] I would like to generate all the combinations of this list while keeping the order of the inner lists unchanged, so [[1, 1, 2], 0, 0, [1, 1, 1, 2], [1, 1, 2], 0] is fine, but [[1, 1, 1, 2], [1, 1, 2], 0, 0, [1, 1, 2], 0] isn't. I've got the feeling that this should be fairly easy in Python, but I just don't see it. Could somebody help me out?

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• Loop through different sets of unique permutations

  - by user558610

  Hi I'm having a hard time getting started to layout code for this problem. I have a fixed amount of random numbers, in this case 8 numbers. R[] = { 1, 2, 3, 4, 5, 6, 7, 8 }; That are going to be placed in 3 sets of numbers, with the only constraint that each set contain minimum one value, and each value can only be used once. For example: R1[] = { 1, 4 } R2[] = { 2, 8, 5, 6 } R3[] = { 7, 4 } I need to loop through all possible combinations of a set R1, R2, R3. Order is not important, so if the above example happened, I don't need R1[] = { 4, 1 } R2[] = { 2, 8, 5, 6 } R3[] = { 7, 4 } NOR R1[] = { 2, 8, 5, 6 } R2[] = { 7, 4 } R3[] = { 1, 4 } What is a good method?

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• How to calculate the cycles that change one permutation into another?

  - by fortran

  Hi, I'm looking for an algorithm that given two permutations of a sequence (e.g. [2, 3, 1, 4] and [4, 1, 3, 2]) calculates the cycles that are needed to convert the first into the second (for the example, [[0, 3], [1, 2]]). The link from mathworld says that Mathematica's ToCycle function does that, but sadly I don't have any Mathematica license at hand... I'd gladly receive any pointer to an implementation of the algorithm in any FOSS language or mathematics package. Thanks!

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• Correct permutation cycle for Verhoeff algorithm

  - by James

  Hello, I'm implementing the Verhoeff algorithm for a check digit scheme, but there seems to be some disagreement in web sources as to which permutation cycle should form the basis of the permutation table. Wikipedia uses: (36)(01589427) while apparently, Numerical Recipies uses a different cycle and this book uses: (0)(14)(23)(56789), quoted from a 1990 article by Winters. It also notes that Verhoeff used the one Wikipedia quotes. Now, my number theory is a little rusty, but the Wikipedia cycle clearly will repeat after the 8th power, while the book one will take 10, despite it saying that s^8=s. Table 2.14(b) has other errors in the 2-cycles, so this is dubious anyway. Unfortunately, I don't have copies of the original articles (and am too tight to pay/disgusted that 40-year old knowledge is still being held to ransom by publishers), nor a copy of Numerical Recipes to check (and am loath to install their paranoia-induced copy protection plug-in to view online). So does any one know which is correct? Are they both correct?

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• Doing permutation of different arrays in perl

  - by nubie2

  Hello! I want to do permutation in perl. For example I have three arrays. ["big", "tiny", "small"] and then I have ["red", "yellow", "green"] and also ["apple", "pear", "banana"]. How do I get: ["big", "red", "apple"] ["big", "red", "pear"] ..etc.. ["small", "green", "banana"] I understand this is called permutation. But I am not sure how to do it. Also I don't know how many arrays I can have. There may be three or four, so I don't want to do nested loop.

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• How would you calculate all possible permutations of 0 through N iteratively?

  - by Bob Aman

  I need to calculate permutations iteratively. The method signature looks like: int[][] permute(int n) For n = 3 for example, the return value would be: [[0,1,2], [0,2,1], [1,0,2], [1,2,0], [2,0,1], [2,1,0]] How would you go about doing this iteratively in the most efficient way possible? I can do this recursively, but I'm interested in seeing lots of alternate ways to doing it iteratively.

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• Is there an expand.grid like function in R, returning permutations ?

  - by Brani

  To become more specific, here is an example: > expand.grid(5, 5, c(1:4,6),c(1:4,6)) Var1 Var2 Var3 Var4 1 5 5 1 1 2 5 5 2 1 3 5 5 3 1 4 5 5 4 1 5 5 5 6 1 6 5 5 1 2 7 5 5 2 2 8 5 5 3 2 9 5 5 4 2 10 5 5 6 2 11 5 5 1 3 12 5 5 2 3 13 5 5 3 3 14 5 5 4 3 15 5 5 6 3 16 5 5 1 4 17 5 5 2 4 18 5 5 3 4 19 5 5 4 4 20 5 5 6 4 21 5 5 1 6 22 5 5 2 6 23 5 5 3 6 24 5 5 4 6 25 5 5 6 6 This data frame was created from all combinations of the supplied vectors. I would like to create a similar data frame from all permutations of the supplied vectors. Notice that each row must contain exactly 2 fives, yet not necessarily as the fist two elements. Thank you.

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• Algorithm to generate all possible permutations of a list?

  - by LLer

  Say I have a list of n elements, I know there are n! possible ways to order these elements. What is an algorithm to generate all possible orderings of this list? Example, I have list [a, b, c]. The algorithm would return [[a, b, c], [a, c, b,], [b, a, c], [b, c, a], [c, a, b], [c, b, a]]. I'm reading this here http://en.wikipedia.org/wiki/Permutation#Algorithms_to_generate_permutations But Wikipedia has never been good at explaining. I don't understand much of it.

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• Data clean up: are there libraries of common permutations that we can use? Or is there a better appr

  - by anyaelena

  We are working on clean-up and analysis of a lot of human-entered customer data. We need to decide programmatically whether 2 addresses (for example) are the same, even though the data was entered with slight variations. Right now we run each address through fairly simplistic string replacement (replacing avenue with ave, for example), concatenate the fields and compare the results. We are doing something similar with names. At the very least, it seems like our list of search-replace values should already exist somewhere. Or perhaps you can suggest a totally different and superior way to detect matches?

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• How to generate all permutations of a string in PHP ?

  - by Johan

  I need an algorithm that return all possible combination of all characters in one string. I've tried: $langd = strlen($input); for($i = 0;$i < $langd; $i++){ $tempStrang = NULL; $tempStrang .= substr($input, $i, 1); for($j = $i+1, $k=0; $k < $langd; $k++, $j++){ if($j > $langd) $j = 0; $tempStrang .= substr($input, $j, 1); } $myarray[] = $tempStrang; } But that only returns the same amount combination as the length of the string. Say the $input is = "hey", the result would be: hey, hye, eyh, ehy, yhe, yeh.

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• is it possible to dynamically set the level of for loop nesting

  - by galaxy

  I'm working out an algorithm to get permutations like 123 132 213 231 312 321 I'm doing it using nested foreach loops. for (..) { for(..) { for(..) { echo $i . $j . $k . "<br />"; } } } Problem is those # of nested loops are optimized for 3-spot permutations. How can I could I dynamically set the number of nested for loops to generate 4-letter or 5-letter permutations?

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• How do I tell GWT to not compile a permutation for gears

  - by Clinton Bosch

  I have included gwt-html5-geolocation into my GWT project and was disappointed to find that it doubled up on my number of permutations compiled. Apparently if the browser does not support geolocation API then it falls back to use gears to find out your location. Is there a way to NOT compile a permutation for gears similar to the way you can tell GWT to only compile certain browser permutations? (the geolocation stuff is very much a nice-to-have and frankly if the client is running an old browser then I am happy not to get their location) Thanks

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• Maximal Length of List to Shuffle with Python random.shuffle?

  - by Henrik

  I have a list which I shuffle with the Python built in shuffle function (random.shuffle) However, the Python reference states: Note that for even rather small len(x), the total number of permutations of x is larger than the period of most random number generators; this implies that most permutations of a long sequence can never be generated. Now, I wonder what this "rather small len(x)" means. 100, 1000, 10000,... Can anybody clarify? Thanks!

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• Generating all unique combinations for "drive ya nuts" puzzle

  - by Yuval A

  A while back I wrote a simple python program to brute-force the single solution for the drive ya nuts puzzle. The puzzle consists of 7 hexagons with the numbers 1-6 on them, and all pieces must be aligned so that each number is adjacent to the same number on the next piece. The puzzle has ~1.4G non-unique possibilities: you have 7! options to sort the pieces by order (for example, center=0, top=1, continuing in clockwise order...). After you sorted the pieces, you can rotate each piece in 6 ways (each piece is a hexagon), so you get 6**7 possible rotations for a given permutation of the 7 pieces. Totalling: 7!*(6**7)=~1.4G possibilities. The following python code generates these possible solutions: def rotations(p): for i in range(len(p)): yield p[i:] + p[:i] def permutations(l): if len(l)<=1: yield l else: for perm in permutations(l[1:]): for i in range(len(perm)+1): yield perm[:i] + l[0:1] + perm[i:] def constructs(l): for p in permutations(l): for c in product(*(rotations(x) for x in p)): yield c However, note that the puzzle has only ~0.2G unique possible solutions, as you must divide the total number of possibilities by 6 since each possible solution is equivalent to 5 other solutions (simply rotate the entire puzzle by 1/6 a turn). Is there a better way to generate only the unique possibilities for this puzzle?

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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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• Help me finish this Python 3.x self-challenge.

  - by Hamish Grubijan

  This is not a homework. I saw this article praising Linq library and how great it is for doing combinatorics stuff, and I thought to myself: Python can do it in a more readable fashion. After half hour of dabbing with Python I failed. Please finish where I left off. Also, do it in the most Pythonic and efficient way possible please. from itertools import permutations from operator import mul from functools import reduce glob_lst = [] def divisible(n): return (sum(j*10^i for i,j in enumerate(reversed(glob_lst))) % n == 0) oneToNine = list(range(1, 10)) twoToNine = oneToNine[1:] for perm in permutations(oneToNine, 9): for n in twoToNine: glob_lst = perm[1:n] #print(glob_lst) if not divisible(n): continue else: # Is invoked if the loop succeeds # So, we found the number print(perm) Thanks!

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• Help me finish this Python self-challenge.

  - by Hamish Grubijan

  This is not a homework. I saw this article praising Linq library and how great it is for doing combinatorics stuff, and I thought to myself: Python can do it in a more readable fashion. After half hour of dabbing with Python I failed. Please finish where I left off. Also, do it in the most Pythonic and efficient way possible please. from itertools import permutations from operator import mul from functools import reduce glob_lst = [] def divisible(n): return (sum(j*10^i for i,j in enumerate(reversed(glob_lst))) % n == 0) oneToNine = list(range(1, 10)) twoToNine = oneToNine[1:] for perm in permutations(oneToNine, 9): for n in twoToNine: glob_lst = perm[1:n] #print(glob_lst) if not divisible(n): continue else: # Is invoked if the loop succeeds # So, we found the number print(perm) Thanks!

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• Fast permutation -> number -> permutation mapping algorithms

  - by ijw

  I have n elements. For the sake of an example, let's say, 7 elements, 1234567. I know there are 7! = 5040 permutations possible of these 7 elements. I want a fast algorithm comprising two functions: f(number) maps a number between 0 and 5039 to a unique permutation, and f'(permutation) maps the permutation back to the number that it was generated from. I don't care about the correspondence between number and permutation, providing each permutation has its own unique number. So, for instance, I might have functions where f(0) = '1234567' f'('1234567') = 0 The fastest algorithm that comes to mind is to enumerate all permutations and create a lookup table in both directions, so that, once the tables are created, f(0) would be O(1) and f('1234567') would be a lookup on a string. However, this is memory hungry, particularly when n becomes large. Can anyone propose another algorithm that would work quickly and without the memory disadvantage?

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• Efficient algorithm to generate all solutions of a linear diophantine equation with ai=1

  - by Ben

  I am trying to generate all the solutions for the following equations for a given H. With H=4 : 1) ALL solutions for x_1 + x_2 + x_3 + x_4 =4 2) ALL solutions for x_1 + x_2 + x_3 = 4 3) ALL solutions for x_1 + x_2 = 4 4) ALL solutions for x_1 =4 For my problem, there are always 4 equations to solve (independently from the others). There are a total of 2^(H-1) solutions. For the previous one, here are the solutions : 1) 1 1 1 1 2) 1 1 2 and 1 2 1 and 2 1 1 3) 1 3 and 3 1 and 2 2 4) 4 Here is an R algorithm which solve the problem. library(gtools) H<-4 solutions<-NULL for(i in seq(H)) { res<-permutations(H-i+1,i,repeats.allowed=T) resum<-apply(res,1,sum) id<-which(resum==H) print(paste("solutions with ",i," variables",sep="")) print(res[id,]) } However, this algorithm makes more calculations than needed. I am sure it is possible to go faster. By that, I mean not generating the permutations for which the sums is H Any idea of a better algorithm for a given H ?

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• Depth First Search Basics

  - by cam

  I'm trying to improve my current algorithm for the 8 Queens problem, and this is the first time I'm really dealing with algorithm design/algorithms. I want to implement a depth-first search combined with a permutation of the different Y values described here: http://en.wikipedia.org/wiki/Eight_queens_puzzle#The_eight_queens_puzzle_as_an_exercise_in_algorithm_design I've implemented the permutation part to solve the problem, but I'm having a little trouble wrapping my mind around the depth-first search. It is described as a way of traversing a tree/graph, but does it generate the tree graph? It seems the only way that this method would be more efficient only if the depth-first search generates the tree structure to be traversed, by implementing some logic to only generate certain parts of the tree. So essentially, I would have to create an algorithm that generated a pruned tree of lexigraphic permutations. I know how to implement the pruning logic, but I'm just not sure how to tie it in with the permutation generator since I've been using next_permutation. Is there any resources that could help me with the basics of depth first searches or creating lexigraphic permutations in tree form?

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