combinatorics

问题 So I am stuck with this problem of trying to find all k-elements subsets from a given N-elements set. I know what the total number of k-subsets is using the formula C(n,k)=C(n-1, k-1)+C(n-1, k) and I also have an idea how to do it in a iterative manner, but I get stuck when I try to think of a recursive solution. Can anyone give me a hint? Thanks! 回答1: For each element of the set, take that element, then add in turn to that all (k-1) subsets of the remaining N-1 element set. "It was a dark

问题 Given 2 arrays Array1 = {a,b,c...n} and Array2 = {10,20,15....x} how can I generate all possible combination as Strings a(i) b(j) c(k) n(p) where 1 <= i <= 10, 1 <= j <= 20 , 1 <= k <= 15, .... 1 <= p <= x Such as: a1 b1 c1 .... n1 a1 b1 c1..... n2 ...... ...... a10 b20 c15 nx (last combination) So in all total number of combination = product of elements of array2 = (10 X 20 X 15 X ..X x) Similar to a Cartesian product, in which the second array defines the upper limit for each element in

问题 Given n sorted lists A1, A2, ..., An of integers in decreasing order, is there an algorithm to efficiently generate all elements of their cartesian product in decreasing tuple sum order ? For example, n=3 A1 = [9, 8, 0] A2 = [4, 2] A3 = [5, 1] The expected output would be the cartesian product of A1xA2xA3 in the following order: combination sum 9, 4, 5 18 8, 4, 5 17 9, 2, 5 16 8, 2, 5 15 9, 4, 1 14 8, 4, 1 13 9, 2, 1 12 8, 2, 1 11 0, 4, 5 9 0, 2, 5 7 0, 4, 1 5 0, 2, 1 3 回答1: If the problem

问题 A k-ary necklace of length n is an ordered list of length n whose items are drawn from an alphabet of length k, which is the lexicographically first list in a sort of all lists sharing an ordering under rotation. Example: (1 2 3) and (1 3 2) are the necklaces of length 3 from the alphabet {1 2 3}. More info: http://en.wikipedia.org/wiki/Necklace_(combinatorics) I'd like to generate these in Scheme (or a Lisp of your choice). I've found some papers... Savage - A New Algorithm for Generating

问题 I am trying to solve a problem where I have pairs like: A C B F A D D C F E E B A B B C E D F D and I need to group them in groups of 3 where I must have a triangule of matching from that list. Basically I need a result if its possible or not to group a collection. So the possible groups are ( ACD and BFE ), or ( ABC and DEF ) and this collection is groupable since all letters can be grouped in groups of 3 and no one is left out. I made a script where I can achieve this for small ammounts of

问题 This question already has answers here : Closed 8 years ago . Possible Duplicate: How to iteratively generate k elements subsets from a set of size n in java? I want to build my own poker hand evaluator but am having trouble with a particular part. If two players get dealt a two card hand, then there will be 48 cards left in the pack. In Texas Hold'em a further 5 possible community cards are then dealt (this is called the board). I want to enumerate / loop through all the 48 choose 5 possible

问题 I want to iterate the entries above the main diagonal of a square matrix of size n without using nested loops. For example if the matrix M is of size n=3, then there are choose(3,2)=3 entries above the diagonal. where choose is the binomial coefficient. for(i=1 to choose(n,2)) row = getRow(i) col = getCol(i) M[row,col] = some value I have not been able to come up with a formula for getting row and col based on index i. For example: for a matrix of size = 3 and indexes starting at 1, i=1

问题 I have two-level nested list like following. [[0, 1], [2], [3, 4], [5, 5], [6], [7], [8], [9], [10, 11], [12]] I want to generate all 8 unique permutations of this nested list, but my application absolutely needs the output to be (pseudo-)randomized and unordered. Usually, permutation strategies produce the permutations in order, but I want to be able to produce all permutations out of order. Moreover, this MUST be done through some generator, as the nested list can be very long, and number

问题 I want to perform a variation of a "6 choose 2" operation. I have written the following code. public void choosePatterns(){ String[] data = {"1", "2", "3", "4", "5", "6"}; String[] originalPattern = new String[15]; int index = 0; for(int i = 0; i < (6-1); i++) { for(int j = i+1; j < 6; j++) { System.out.println(i + "," +j); } } } My code so far generates all the possible combinations of "6 choose 2." However, I would like to vary this and print all the remaining numbers. So far instance, if

问题 I'm trying to achieve a loop over all possible 2 letter combinations. Something like foreach(range(aa,zz) as $i) {...} My current solution is: foreach (range(a, z) as $first) { foreach (range(a, z) as $second) { //all 2 letter combinations echo $first.$second; } } This makes me worry that if I needed all possible 10 letter combinations, there would be 10 loops involved. Is there a better way to achieve this? 回答1: You could loop over letters using a simple for loop : for ($letter = 'aa';