Here is my version, which works purely in integers (the division by k always produces an integer quotient) and is fast at O(k): function choose(n, k) if k == 0 return 1 return (n * choose(n – 1, k – 1)) / k I wrote it recursively because it’s so simple and pretty, but you … Read more
OK, here’s your code (and btw, thanks for posting such an interesting and challenging problem – at least for me… :-)) – using recursion for all possible permutations (by N) given an array of elements) Code : <?php function permutations($arr,$n) { $res = array(); foreach ($arr as $w) { if ($n==1) $res[] = $w; else … Read more
Here is a conceptual answer and a code based on how lex ordering works. (So I guess my answer is like that of “moron”, except that I think that he has too few details and his links have too many.) I wrote a function unchoose(n,S) for you that works assuming that S is an ordered … Read more
Almost a Base Conversion This solution is motivated by the observation that, if it were not for the possibility of repeating the character at array index 0 in the high-order position of a valid combination, this problem would simply be a base conversion from decimal to the new base of all integers from 0 to … Read more
This is a well-studied problem of generating all k-subsets, or k-combinations, which can be easily done without recursion. The idea is to have array of size k keeping sequence of indices of elements from the input array (which are numbers from 0 to n – 1) in increasing order. (Subset then can be created by … Read more
How about using outer? But this particular function concatenates them into one character string. outer( c(“aa”, “ab”, “cc”), c(“aa”, “ab”, “cc”) , “paste” ) # [,1] [,2] [,3] #[1,] “aa aa” “aa ab” “aa cc” #[2,] “ab aa” “ab ab” “ab cc” #[3,] “cc aa” “cc ab” “cc cc” You can also use combn on … Read more
I think what you’re looking for is np.mgrid. Unfortunately, this returns the array in a format that’s different from what you need, so you’ll need to do a little post-processing: a = np.mgrid[0:4, 0:4, 0:11] # All points in a 3D grid within the given ranges a = np.rollaxis(a, 0, 4) # Make the 0th … Read more
Since I offered an ODBC approach I thought I should elaborate on it, as it is not immediately obvious how to do this. And, in honesty, I needed to relearn the process and document it for myself. This is a way to generate a Cartesian product of two or more one-dimensional data arrays using Excel … Read more
Use recursion. int findCombinationsCount(int amount, int coins[]) { return findCombinationsCount(amount, coins, 0); } int findCombinationsCount(int amount, int coins[], int checkFromIndex) { if (amount == 0) return 1; else if (amount < 0 || coins.length == checkFromIndex) return 0; else { int withFirstCoin = findCombinationsCount(amount-coins[checkFromIndex], coins, checkFromIndex); int withoutFirstCoin = findCombinationsCount(amount, coins, checkFromIndex+1); return withFirstCoin + … Read more
How many combinations are there? So first the question how many combinations are there? And the answer is you have to multiply the amount of every array with each other. So (c = amount1): carray 1 * carray 2 * … * carray n And specific for your example: carray 1 * carray 2 * … Read more