combinatorics

问题 I know that there is an algorithm that permits, given a combination of number (no repetitions, no order), calculates the index of the lexicographic order. It would be very useful for my application to speedup things... For example: combination(10, 5) 1 - 1 2 3 4 5 2 - 1 2 3 4 6 3 - 1 2 3 4 7 .... 251 - 5 7 8 9 10 252 - 6 7 8 9 10 I need that the algorithm returns the index of the given combination. es: index( 2, 5, 7, 8, 10 ) --> index EDIT: actually I\'m using a java application that

问题 A multi-set is a set in which all the elements may not be unique.How to enumerate all the possible permutations among the set elements? 回答1: Generating all the possible permutations and then discarding the repeated ones is highly inefficient. Various algorithms exist to directly generate the permutations of a multiset in lexicographical order or other kind of ordering. Takaoka's algorithm is a good example, but probably that of Aaron Williams is better http://webhome.csc.uvic.ca/~haron

问题 Here I try to write a program in C++ to find NCR. But I\'ve got a problem in the result. It is not correct. Can you help me find what the mistake is in the program? #include <iostream> using namespace std; int fact(int n){ if(n==0) return 1; if (n>0) return n*fact(n-1); }; int NCR(int n,int r){ if(n==r) return 1; if (r==0&&n!=0) return 1; else return (n*fact(n-1))/fact(n-1)*fact(n-r); }; int main(){ int n; //cout<<\"Enter A Digit for n\"; cin>>n; int r; //cout<<\"Enter A Digit for r\"; cin>>r

问题 I\'m looking for an efficient way to achieve this: you have a list of numbers 1.....n (typically: 1..5 or 1..7 or so - reasonably small, but can vary from case to case) you need all combinations of all lengths for those numbers, e.g. all combinations of just one number ({1}, {2}, .... {n}), then all combinations of two distinct numbers ({1,2}, {1,3}, {1,4} ..... {n-1, n} ), then all combinations fo three of those numbers ({1,2,3}, {1,2,4}) and so forth Basically, within the group, the order

问题 For a set of the form A = {1, 2, 3, ..., n} . It is called partition of the set A , a set of k<=n elements which respect the following theorems: a) the union of all the partitions of A is A b) the intersection of 2 partitions of A is the empty set (they can\'t share the same elements). For example. A = {1, 2,... n} We have the partitions: {1, 2, 3} {1, 2} {3} {1, 3} {2} {2, 3} {1} {1} {2} {3} These theoretical concepts are presented in my algorithms textbook (by the way, this subchapter is

问题 I\'m trying to write some code to test out the Cartesian product of a bunch of input parameters. I\'ve looked at itertools , but its product function is not exactly what I want. Is there a simple obvious way to take a dictionary with an arbitrary number of keys and an arbitrary number of elements in each value, and then yield a dictionary with the next permutation? Input: options = {\"number\": [1,2,3], \"color\": [\"orange\",\"blue\"] } print list( my_product(options) ) Example output: [ {\

问题 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

问题 What is the most efficient way to generate all the combinations, dispositions and permutations of an array in PHP? 回答1: Here is code to get all permutations: http://php.net/manual/en/function.shuffle.php#90615 With the code to get the power set, permutations are those of maximal length, the power set should be all combinations. I have no idea what dispositions are, so if you can explain them, that would help. 回答2: You can use this class: http://pear.php.net/package/Math_Combinatorics and use

问题 When we sort a list, like a = [1,2,3,3,2,2,1] sorted(a) => [1, 1, 2, 2, 2, 3, 3] equal elements are always adjacent in the resulting list. How can I achieve the opposite task - shuffle the list so that equal elements are never (or as seldom as possible) adjacent? For example, for the above list one of the possible solutions is p = [1,3,2,3,2,1,2] More formally, given a list a , generate a permutation p of it that minimizes the number of pairs p[i]==p[i+1] . Since the lists are large,

问题 There are n people numbered from 1 to n . I have to write a code which produces and print all different combinations of k people from these n . Please explain the algorithm used for that. 回答1: I assume you're asking about combinations in combinatorial sense (that is, order of elements doesn't matter, so [1 2 3] is the same as [2 1 3] ). The idea is pretty simple then, if you understand induction / recursion: to get all K -element combinations, you first pick initial element of a combination